{"id":28717,"date":"2022-09-30T17:52:30","date_gmt":"2022-09-30T12:22:30","guid":{"rendered":"https:\/\/apboardsolutions.com\/?p=28717"},"modified":"2023-03-30T14:32:15","modified_gmt":"2023-03-30T09:02:15","slug":"ap-inter-1st-year-physics-study-material-chapter-8","status":"publish","type":"post","link":"https:\/\/apboardsolutions.com\/ap-inter-1st-year-physics-study-material-chapter-8\/","title":{"rendered":"AP Inter 1st Year Physics Study Material Chapter 8 Oscillations"},"content":{"rendered":"

Andhra Pradesh BIEAP\u00a0AP Inter 1st Year Physics Study Material<\/a> 8th Lesson Oscillations Textbook Questions and Answers.<\/p>\n

AP Inter 1st Year Physics Study Material 8th Lesson Oscillations<\/h2>\n

Very Short Answer Questions<\/span><\/p>\n

Question 1.
\nGive two examples of periodic motion which are not oscillatory.
\nAnswer:<\/p>\n

    \n
  1. The motion of planets around the sun.<\/li>\n
  2. The motion of an electron round the nucleus.<\/li>\n<\/ol>\n

    Question 2.
    \nThe displacement in S.H.M. is given by y = a sin (20t + 4). What is the displacement when it is increased by 2\u03c0\/\u03c9 ?
    \nAnswer:
    \nThe displacement in S.H.M. is y = a sin (20t + 4)
    \nThe time period T = \\(\\frac{2 \\pi}{\\omega}\\) is increased, the displacement of the particle remains the same.<\/p>\n

    \"AP<\/p>\n

    Question 3.
    \nA girl is swinging seated in a swing. What is the effect on the frequency of oscillation if she stands ?
    \nAnswer:
    \nFrequency (n) = \\(\\frac{1}{2 \\pi} \\sqrt{\\frac{g}{l}}\\), n \u221d \\(\\frac{1}{\\sqrt{l}}\\)
    \nA girl swinging in standing position location of centre of mass shifts upwards l decreases, frequency of oscillation increases.<\/p>\n

    Question 4.
    \nThe bob of a simple pendulum is a hollow sphere filled with water. How will the period of oscillation change, if the water begins to drain out of the hollow sphere ?
    \nAnswer:
    \nThe period of the pendulum is same, when the bob is hollow (or) completely filled with water. As water flows out from the bob, the centre of gravity of the bob lowered. The pendulum length increases. Hence time period also increases. When the bob becomes empty, again centre of gravity shifts upwards. The pendulum length decreases. The time period also decreases.<\/p>\n

    Question 5.
    \nThe bob of a simple pendulum is made of wood. What will be the effect on the time period if the wooden bob is replaced by an identical bob of aluminum ?
    \nAnswer:
    \nTime period (T) = \\(2 \\pi \\sqrt{\\frac{l}{g}}\\), Time period is independent of mass of the bob.
    \nHence wooden bob is replaced by an identical aluminium bob, Time period remains constant.<\/p>\n

    Question 6.
    \nWill a pendulum clock gain or lose time when taken to the top of a mountain.?
    \nAnswer:
    \nT \u221d \\(\\frac{1}{\\sqrt{g}}\\), At the mountain top, the value of g decreases, hence time period increases i.e. the pendulum will take longer time to complete one vibration. Hence pendulum clock will loose time on the mountain top.<\/p>\n

    Question 7.
    \nA pendulum dock gives correct time at the equator. Will it gain or lose time if it is taken to the poles ? If so, why ?
    \nAnswer:
    \nTime period (T) = \\(2 \\pi \\sqrt{\\frac{l}{g}}\\)
    \ng value at poles is greater than at equator.
    \nIf it is taken to the poles g value increases, time period decreases.
    \nSo pendulum clock gains time.<\/p>\n

    \"AP<\/p>\n

    Question 8.
    \nWhat fraction of the total energy is K.E when the displacement is one half of a amplitude of a particle executing S.H.M.
    \nAnswer:
    \nTotal energy (E) = \\(\\frac{1}{2}\\) m\u03c92<\/sup> A2<\/sup>
    \nGiven y = \\(\\frac{\\mathrm{A}}{2}\\), Kinetic energy = \\(\\frac{1}{2}\\)m\u03c92<\/sup>(A2<\/sup> – y2<\/sup>) = \\(\\frac{1}{2}\\) m\u03c92<\/sup>\\(\\left(A^2-\\frac{A^2}{4}\\right)\\) = \\(\\frac{3}{4}\\) \u00d7 \\(\\frac{1}{2}\\)m\u03c92<\/sup>A2<\/sup>
    \nK.E = \\(\\frac{3}{4}\\) \u00d7 E
    \n\u2234 \\(\\frac{K \\cdot E}{E}\\) = \\(\\frac{3}{4}\\)<\/p>\n

    Question 9.
    \nWhat happens to the energy of a simple harmonic oscillator if its amplitude is doubled ?
    \nAnswer:
    \nTotal energy (E) = \\(\\frac{1}{2}\\)m\u03c92<\/sup>A2<\/sup>
    \nGiven Amplitude A is doubled
    \nE’ = \\(\\frac{1}{2}\\)m\u03c92<\/sup>(2A)2<\/sup>
    \nE’ = 4 \u00d7 \\(\\frac{1}{2}\\)m\u03c92<\/sup>A2<\/sup>
    \nE’ = 4E
    \n\u2234 Energy becomes four times.<\/p>\n

    Question 10.
    \nCan a simple pendulum be used in an artificial satellite? (T.S. Mar. 16)
    \nAnswer:
    \nNo, in an artificial satellite acceleration due to gravity is zero. Hence we cannot use simple pendulum in an artificial satellite.<\/p>\n

    Short Answer Questions<\/span><\/p>\n

    Question 1.
    \nDefine simple harmonic motion. Give two examples.
    \nAnswer:
    \nSimple harmonic motion : “A body is said to be in simple harmonic motion, if it moves to and fro along a straight line, about its mean position such that, at any point its acceleration is proportional to its displacement but opposite in direction and is directed always towards the mean position”.
    \nw \u221d -x<\/p>\n

    If a is the acceleration of the body at any given displacement x from the mean position, the time for the body to be in S.H.M.
    \nDisplacement of a particle in S.H.M is given by
    \nx(t) = A cos (\u03c9t + \u03d5)
    \n\"AP<\/p>\n

    Example:<\/p>\n

      \n
    1. Motion of a simple pendulum.<\/li>\n
    2. Motion of mass attached to a spring.<\/li>\n
    3. Motion of atoms in solids.<\/li>\n
    4. Cork floating on water.<\/li>\n<\/ol>\n

      \"AP<\/p>\n

      Question 2.
      \nPresent graphically the variations of displacement, velocity and acceleration with time for a particle in S.H.M.
      \nAnswer:
      \nLet us put \u03d5 = 0 and write the expressions for x(t),
      \n\u03c5(t) and a(t).
      \nx(t) = A cos \u03c9t. \u03c5(t) = -A\u03c9sin\u03c9t
      \na(t) = -\u03c92<\/sup>A cos \u03c9t. The corresponding plots are shown in figure. All quantities varies sinusoidally with time.
      \n\"AP
      \nx(t) varies between – A to A; \u03c5(t) varies from -\u03c9A to \u03c9A and a(t) varies from – \u03c92<\/sup>A to \u03c92<\/sup>A with respect to displacement plot, velocity plot has a phase difference of \\(\\frac{\\pi}{2}\\) and acceleration plot has a phase difference of \u03c0.<\/p>\n

      Question 3.
      \nWhat is phase ? Discuss the phase relations between displacement, velocity and acceleration in simple harmonic motion.
      \nAnswer:
      \nPhase : The phase of a particle executing S.H.M. at any instant is defined as its state (or) condition as regards to its position and direction of motion at that instant.<\/p>\n

        \n
      1. Displacement : x = A cos (\u03c9t – \u03d5), (\u03c9t – \u03d5) is called phase and \u03d5 is epoch.<\/li>\n
      2. Velocity : V = -A\u03c9 sin (\u03c9t – \u03d5), Here also (\u03c9t – \u03d5) is phase angle<\/li>\n
      3. Acceleration : a = -\u03c92<\/sup>A cos (\u03c9t – \u03d5), Here also (\u03c9t – \u03d5) is phase angle.<\/li>\n<\/ol>\n

        Phase difference between displacement and velocity = \\(\\frac{\\pi}{2}\\)
        \nPhase difference between velocity and acceleration = \\(\\frac{\\pi}{2}\\)
        \nPhase difference between displacement and acceleration = \u03c0.<\/p>\n

        Question 4.
        \nObtain an equation for the frequency of oscillation of spring of force constant k to which a mass m is attached. Answer:
        \nLet us consider a spring suspended vertically from a rigid support and loaded with a mass m. If it is now pulled down and released, it executes vertical oscillations about mean position.
        \nRestoring force is directly proportional to the displacement, but oppositely directed
        \n\"AP
        \nF \u221d -y
        \nF = -ky —–> (1)
        \nWhere k is force constant
        \nMa = -ky (\u2234 F = Ma)
        \na = \\(-\\left(\\frac{K}{M}\\right) y\\) —-> (2)
        \nSince K and M are constant. We can write a \u221d -y
        \nie Acceleration is directly proportional to the displacement but oppositely directed. Hence oscillations of a loaded spring is S.H.M.
        \nComparing eq. (2) with a = -\u03c92<\/sup>y
        \n\u03c92<\/sup> = \\(\\frac{K}{M}\\) ; \u03c9 = \\(\\sqrt{\\frac{\\mathrm{K}}{\\mathrm{M}}}\\)
        \nT = \\(\\frac{2 \\pi}{\\omega}\\) ; T = \\(2 \\pi \\sqrt{\\frac{\\mathrm{M}}{\\mathrm{K}}}\\)
        \nFrequency of oscillation (n) = \\(\\frac{1}{2 \\pi} \\sqrt{\\frac{K}{M}}\\) —> (3)<\/p>\n

        Question 5.
        \nDerive expressions for the kinetic energy and potential energy of a simple harmonic oscillator.
        \nAnswer:
        \nKinetic energy of simple harmonic oscillator : The velocity of a particle in S.H.M. is given by
        \nv = \\(\\omega \\sqrt{A^2-y^2}\\) \u2234 Kinetic energy = \\(\\frac{1}{2}\\) mv2<\/sup> = \\(\\frac{1}{2}\\) m\u03c92<\/sup>(A2<\/sup> – y2<\/sup>)
        \nWhen y = 0, (K.E)Max<\/sub> = \\(\\frac{1}{2}\\) m\u03c92<\/sup>A2<\/sup> (Mean position)
        \nWhen y = A, (K.E)Min<\/sub> = 0 (Extreme position)
        \nK.E is maximum at mean position and minimum at extreme position.<\/p>\n

        Potential energy of simple harmonic oscillator: When the displacement of a particle executing simple Harmonic oscillations increases, the restoring force also increases. The restoring force is in the opposite direction to the displacement. Therefore work is done in moving through the displacement, against restoring force. If F is the restoring force at the displacement y.
        \nThe average force against which work is done = \\(\\frac{\\mathrm{O}+\\mathrm{F}}{2}\\) = \\(\\frac{F}{2}\\)<\/p>\n

        \u2234 Workdone on the particle for the displacement (y) = Average force \u00d7 displacement F
        \ni.e. w = \\(\\frac{\\mathrm{F}}{2}\\) \u00d7 y
        \nw = \\(\\frac{\\max y}{2}\\) —- (1) (\u2235 F = ma)
        \nBut acceleration of a particle in S.H.M. is given by
        \na = \u03c92<\/sup>y —– (2)
        \nUsing eq\u2019s (1) and (2). we get
        \nThe work done (w) = \\(\\frac{1}{2} m \\omega^2 y^2\\) mw2y2
        \nThis work done is stored in the form of P.E.
        \n\u2234 P.E. = \\(\\frac{1}{2} m \\omega^2 y^2\\) —– (3)
        \nIf y = 0, (P.E)Min<\/sub> = 0 (At mean position)
        \nIf y = A, (P.E)Max<\/sub> = \\(\\frac{1}{2} m \\omega^2 A^2\\) (At extreme position).
        \n\u2234 P.E. is maximum at extreme position and minimum at mean position.<\/p>\n

        Question 6.
        \nHow does the energy of a simple pendulum vary as it moves from one extreme position to the other during its oscillations?
        \nAnswer:
        \nThe total energy associated with a particle executing S.H.M. at any point is the sum of potential energy and kinetic energy at that point.
        \n\"AP
        \nTotal energy (E) = K.E + P.E
        \nK.E = \\(\\frac{1}{2} m \\omega^2\\left(A^2-y^2\\right)\\)
        \nP.E. = \\(\\frac{1}{2} m \\omega^2 y^2\\)
        \nT.E. = \\(\\frac{1}{2}\\)m\u03c92<\/sup>(A2<\/sup> – y2<\/sup>) + \\(\\frac{1}{2}\\)m\u03c92<\/sup>y2<\/sup> = \\(\\frac{1}{2}\\)m\u03c92<\/sup>A2<\/sup>
        \nAt mean position y = 0, P.E = 0
        \nand (K.E)Max<\/sub> = \\(\\frac{1}{2} m \\omega^2 A^2\\)
        \n\u2234 T.E. = 0 + \\(\\frac{1}{2} m \\omega^2 A^2\\) = \\(\\frac{1}{2} m \\omega^2 A^2\\)
        \nAt extreme position y = A, K.E = 0 and P.E = \\(\\frac{1}{2} m \\omega^2 A^2\\)
        \n\u2234 (P.E)Max<\/sub> = \\(\\frac{1}{2} m \\omega^2 A^2\\)
        \nFrom mean position to extreme position K.E. is to be converted into P.E.<\/p>\n

        \"AP<\/p>\n

        Question 7.
        \nDerive the expressions for displacement, velocity and acceleration of a particle executes S.H.M.
        \nAnswer:
        \nConsider a particle P moves on the circumference of a circle of radius A with uniform angular velocity \u03c9. Let PN be the perpendicular drawn to the diameter yy’ from P.
        \nAs P moves on the circumference of the circle, N moves on the diameter yy’ to and fro about the centre O.
        \n\"AP
        \n\"AP<\/p>\n

        Velocity : The velocity of a partcile executing SHM is given by
        \nv = \\(\\frac{d y}{d t}\\) = \\(\\frac{\\mathrm{d}}{\\mathrm{dt}}\\) (A sin \u03c9t)
        \n= A\u03c9 cos \u03c9t = A\u03c9\\(\\sqrt{1-\\sin ^2 \\omega t}\\)
        \nv = \\(A \\omega \\sqrt{1-\\left(\\frac{y}{A}\\right)^2}\\) (\u2235 sin \u03c9t = \\(\\frac{y}{A}\\))
        \nv = \u03c9\\(\\omega \\sqrt{A^2-y^2}\\) —- (2)<\/p>\n

        Acceleration : As the rate of change of velocity gives acceleration of the particle executing S.H.M is given by
        \na = \\(\\frac{d v}{d t}\\) = \\(\\frac{\\mathrm{d}}{\\mathrm{dt}}\\)(A\u03c9 cos \u03c9t) = -A\u03c92<\/sup> sin \u03c9t
        \n\u2234 a = -\u03c92<\/sup>y —- (3)<\/p>\n

        Long Answer Questions<\/span><\/p>\n

        Question 1.
        \nDefine simple harmonic motion. Show that the motion of (point) projection of a particle performing uniform circular motion, on any diameter, is simple harmonic. (AP – Mar. \u201818; TS – Mar. \u201816)
        \nAnswer:
        \nSimple harmonic motion : A body is said to be in simple harmonic motion, if it moves to and fro along a straight line, about its mean position such that, at any point its acceleration is proportional to its displacement but opposite in direction and directed always towards the mean position.<\/p>\n

        Show that the projection of uniform circular motion on any diameter is simple harmonic : Consider a particle P moving on the circumference of a circle of radius A with uniform angular velocity \u03c9. Let O be the centre of the circle. XX’ and YY’ are two mutually perpendicular diameters of the circle as shown in the figure. Let PN be drawn perpendicular to the diameter YY’ from P. As P moves on the circumference of the circle, N moves on the diameter YY’ to and fro about the centre O. Let us consider the position of N at any time t, after leaving the point ‘O’, during its motion. The corresponding angular displacement of the particle P is \"AP
        \n\"AP
        \nHence acceleration is directly proportional to the displacement and opposite direction. Hence motion of N is simple harmonic.<\/p>\n

        Question 2.
        \nShow that the motion of a simple pendulum is simple harmonic and hence derive an equation for its time period. What is seconds pendulum ? (TS – Mar. ’18, ’17, ’15, ’14, ’13; AP – Mar. ’17, ’16, ’15, ’14, ’13)
        \nAnswer:<\/p>\n

          \n
        1. Consider simple pendulum, a small bob of mass m tied to an inextensible mass less string of length L and other end of the string is fixed from a rigid support.<\/li>\n
        2. Once the bob is slightly displaced and released, it begins to oscillate about mean position.<\/li>\n
        3. Let \u03b8 be the angular displacement and T be the tension in the string.<\/li>\n
        4. The forces acting on the bob are (a) tension T along the string (b) weight mg acts vertically downwards.<\/li>\n
        5. The force mg can be resolved into two components (1) mg cos \u03b8 along the PA and (2) mg sin \u03b8 acts along PB.
          \n\"AP<\/li>\n
        6. From figure at point P,
          \nT = mg cos \u03b8 —– (1)<\/li>\n
        7. The force mg sin \u03b8 will provide the restoring torque, which tends to bring the bob back to its mean position O.<\/li>\n
        8. The restoring torque is given by
          \n\\(\\tau\\) = Restoring force \u00d7 \u22a5lar<\/sup> distance
          \n\\(\\tau\\) = -mg sin \u03b8 \u00d7 L —– (2)
          \nHere negative sign shows that the torque acts to reduce \u03b8.
          \nThen sin \u03b8 is replaced by \u03b8 i.e., sin \u03b8 \u2248 \u03b8
          \nx = -mg L \u03b8 —– (3) (\u2235 sin \u03b8 = \u03b8 – \\(\\frac{\\theta^3}{3 !}\\) + \\(\\frac{\\theta^3}{5 !}\\) …..)<\/li>\n
        9. From equation (3), we note that \\(\\tau\\) \u221d \u03b8. and This \\(\\tau\\) will bring the bob back towards its equilibrium position.
          \nSo, if the bob is left free, it will execute angular simple harmonic motion.
          \nComparing equation (3), with the equation \\(\\tau\\) = -k\u03b8, we have Spring factor, k = mgL.<\/li>\n
        10. Here inertia factor = Moment of inertia of the bob about the point of suspension = mL2<\/li>\n
        11. In S.H.M, Time period (T) = \\(2 \\pi \\sqrt{\\frac{\\text { Inertia factor }}{\\text { Spring factor }}}\\)
          \nT = \\(2 \\pi \\sqrt{\\frac{m L^2}{m g L}}\\)
          \nT = \\(2 \\pi \\sqrt{\\frac{L}{g}}\\) — (4)
          \nSeconds pendulum : A pendulum whose time period is 2 seconds is called seconds pendulum.
          \nT = 2 seconds.<\/li>\n<\/ol>\n

          \"AP<\/p>\n

          Question 3.
          \nDerive the equation for the kinetic energy and potential energy of a simple harmonic oscillator and show that the total energy of a particle in simple harmonic motion is constant at any point on its path.
          \nAnswer:
          \nKinetic energy : The velocity of a particle in S.H.M is given by v = \u03c9\\(\\sqrt{A^2-y^2}\\)
          \n\u2234 Kinetic energy = \\(\\frac{1}{2} \\mathrm{mv}^2\\) = \\(\\frac{1}{2} m \\omega^2\\left(A^2-y^2\\right)\\) —— (1)
          \nWe know that y = A sin \u03c9t
          \nK.E = \\(\\frac{1}{2}\\)m\u03c92<\/sup>A2<\/sup>[1 – sin2<\/sup>\u03c9t) —— (2)
          \nWhen y = 0, (K.E)Max<\/sub> = –\\(\\frac{1}{2}\\)m\u03c92<\/sup>A2<\/sup> (Mean position)
          \nWhen y = A, (K.E)Min<\/sub> = 0 (At extreme position)
          \n\u2234 K.E is maximum at mean position and minimum at extreme position.<\/p>\n

          Potential energy : When the displacement of a particle executing simple harmonic oscillations increase, the restoring force also increases. The restoring force is in the opposite direction to the displacement. Therefore work is done in moving through the displacement, against restoring force. If F is restoring force at the displacement y.
          \nThe average force against which work is done = \\(\\frac{\\mathrm{O}+\\mathrm{F}}{2}\\) = \\(\\frac{F}{2}\\)<\/p>\n

          \u2234 Workdone on the particle for the displacement (y) = Average force \u00d7 displacement.
          \ni.e., w = \\(\\frac{1}{2}\\) \u00d7 y
          \nw = \\(\\frac{\\max y}{2}\\) —– (3) (\u2235 F = ma)
          \nBut acceleration of a particle in S.H.M is given by
          \na = -\u03c92<\/sup>y —– (4)
          \nUsing equations (3) and (4), we get
          \nThe work done (W) = \\(\\frac{1}{2}\\)m\u03c92<\/sup>y2<\/sup>
          \nThis work done is stored in the form of P.E
          \n\u2234 P.E = \\(\\frac{1}{2}\\)m\u03c92<\/sup>y2<\/sup> —– (5)
          \n\u2234 P.E = \\(\\frac{1}{2}\\)m\u03c92<\/sup>A2<\/sup> sin2<\/sup>\u03c9t —– (6) (\u2235 y = A sin \u03c9t)
          \nIf y = 0, (P.E)Min<\/sub> = 0 (At mean position)
          \nIf y = A, (P.E)Max<\/sub> = – \\(\\frac{1}{2}\\)m\u03c92<\/sup>A2<\/sup> (At extreme position)
          \n\u2234 P.E. is maximum at extreme position and minimum at mean position.<\/p>\n

          Total energy (E) : The total energy associated with a particle executing S.H.M at any point is the sum of RE and K.E at that point.
          \n\"AP
          \nTotal energy (E) = K.E + P.E
          \nK.E. = \\(\\frac{1}{2}\\)m\u03c92<\/sup>y2<\/sup>
          \n\u2234 T.E. = \\(\\frac{1}{2}\\)m\u03c92<\/sup> (A2<\/sup> – y2<\/sup>) + \\(\\frac{1}{2}\\)m\u03c92<\/sup>y2<\/sup> = \\(\\frac{1}{2}\\)m\u03c92<\/sup>A2<\/sup>
          \nAt mean position y = 0, RE = 0, (KE)Max<\/sub> = – \\(\\frac{1}{2}\\)m\u03c92<\/sup>A2<\/sup>
          \n\u2234 T.E = K.E + P.E
          \nT.E = \\(\\frac{1}{2}\\)m\u03c92<\/sup>A2<\/sup> + 0 = \\(\\frac{1}{2}\\)m\u03c92<\/sup>A2<\/sup>
          \nAt extreme position, y = A, K.E = 0 and,
          \n(P.E)Max<\/sub> = \\(\\frac{1}{2}\\)m\u03c92<\/sup>A2<\/sup>.
          \n\u2234 T.E = K.E + P.E
          \n\"AP
          \nT.E. = O + \\(\\frac{1}{2}\\)m\u03c92<\/sup>A2<\/sup> = \\(\\frac{1}{2}\\)m\u03c92<\/sup>A2<\/sup>
          \nFrom mean position to extreme position K.E is to be converted into RE.<\/p>\n

          Problems<\/span><\/p>\n

          Question 1.
          \nThe bob of a pendulum is made of a hollow brass sphere. What happens to the time period of the pendulum, if the bob is filled with water completely ? Why?
          \nSolution:
          \nTime period (T) = \\(2 \\pi \\sqrt{\\frac{1}{g}}\\)
          \nThe period of the pendulum is same. When the bob is hollow (or) completely filled with water. As water flows out from the bob, the centre of gravity of the bob lowers.<\/p>\n

          The pendulum length increases. Hence time period also increases. When the bob becomes empty, again centre of gravity shifts upwards. The pendulum length decreases. The time period also decreases.<\/p>\n

          \"AP<\/p>\n

          Question 2.
          \nTwo identical springs of force constant “k” are joined one at the end of the other (in series). Find the effective force constant of the combination.
          \nSolution:
          \nk1<\/sub> = k2<\/sub> = k
          \nIf two springs are connected in series
          \n\"AP<\/p>\n

          Question 3.
          \nWhat are the physical quantities having maximum value at the mean position in SHM ?
          \nSolution:<\/p>\n

            \n
          1. Velocity, Vmax<\/sub> = A\u03c9<\/li>\n
          2. Kinetic energy, (K.E)Max<\/sub> = \\(\\frac{1}{2}\\)m\u03c92<\/sup>A2<\/sup>.<\/li>\n<\/ol>\n

            Question 4.
            \nA particle executes SHM such that, the maximum velocity during the oscillation is numerically equal to half the maximum acceleration. What is the time period ?
            \nSolution:
            \nGiven VMax<\/sub> = \\(\\frac{1}{2}\\)aMax<\/sub>
            \nA\u03c9 = \\(\\frac{1}{2}\\)\u03c92<\/sup>
            \n\u03c9 = 2
            \nT = \\(\\frac{2 \\pi}{\\omega}\\) = \\(\\frac{2 \\pi}{2}\\) = \u03c0 sec.<\/p>\n

            Question 5.
            \nA mass of 2 kg attached to a spring of force constant 260 Nm-1<\/sup> makes 100 oscillations. What is the time taken ?
            \nSolution:
            \nm = 2 kg, k = 260 N\/m
            \nT = \\(2 \\pi \\sqrt{\\frac{m}{k}}\\) = 2 \u00d7 3.14\\(\\sqrt{\\frac{2}{260}}\\) = 0.5508sec.
            \n\u2234 Time for 100 oscillations = 100 \u00d7 0.5508
            \n= 55.08 sec.<\/p>\n

            Question 6.
            \nA simple pendulum in a stationery lift has time period T. What would be the effect on the time period when the lift<\/p>\n

            (i) moves up with uniform velocity
            \n(ii) moves down with uniform velocity
            \n(iii) moves up with uniform acceleration a
            \n(iv) moves down with uniform acceleration ‘a’
            \n(v) begins to fall freely under gravity ?<\/p>\n

            Solution:<\/p>\n

            i) When the lift moves up with uniform velocity
            \nT = \\(2 \\pi \\sqrt{\\frac{1}{g}}\\)
            \nNo change in time period.<\/p>\n

            ii) When the lift moves down with uniform velocity. No change in the time period.<\/p>\n

            iii) When the lift moves up with acceleration.
            \nT = \\(2 \\pi \\sqrt{\\frac{1}{g+a}}\\)
            \nTime period decreases.<\/p>\n

            iv) When the lift moves down with acceleration.
            \nT = 2\u03c0\\(\\sqrt{\\frac{1}{g-a}}\\)
            \nTime period increases.<\/p>\n

            v) Lift falls freely, a = g
            \nT = 2\u03c0\\(\\sqrt{\\frac{l}{g-g}}\\) = 2\u03c0\\(\\sqrt{\\frac{1}{0}}\\) = \u221d
            \nTime period becomes infinity.<\/p>\n

            Question 7.
            \nA particle executing SHM has amplitude of 4 cm and its acceleration at a distance of 1 cm from the mean position is 3 cm s-2<\/sup>. What will is velocity be when it is at a distance of 2 cm from its mean position ?
            \nSolution:
            \nA = 4 cm, x1<\/sub> = 1 cm, a = 3 cm\/s2<\/sup>
            \na = \u03c92<\/sup>x1<\/sub>
            \n3 = \u03c92<\/sup> \u00d7 1<\/sub>
            \n\u03c9 = \\(\\sqrt{3}\\)
            \nVelocity v = \u03c9\\(\\sqrt{A^2-x_2^2}\\)
            \n(\u2235 x2 = 2 cm)
            \nv = \\(\\sqrt{3} \\sqrt{4^2-2^2}\\)
            \nv = \\(\\sqrt{3} \\times \\sqrt{12}\\)
            \nv = \\(\\sqrt{36}\\) = 6 cm\/s.<\/p>\n

            Question 8.
            \nA simple harmonic oscillator has a time period of 2s. What will be the change in the phase 0.25 s after leaving the mean position ?
            \nSolution:
            \nT = 2 sec
            \nt = 0.25 sec
            \nsin \u03c9t = sin\\(\\left(\\frac{2 \\pi}{T}\\right) t\\)
            \n\u03d5 = \u03c9t = \\(\\frac{2 \\pi}{T}\\) \u00d7 t
            \n= \\(\\frac{2 \\pi}{2}\\) \u00d7 0.25
            \n\u03d5 = \\(\\frac{\\pi}{4}\\)<\/p>\n

            \"AP<\/p>\n

            Question 9.
            \nA body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. find the acceleration and velocity of the body when the displacement is
            \n(a) 5 cm.
            \n(b) 3 cm.
            \n(c) 0 cm.
            \nSolution:
            \nA = 5 cm = 5 \u00d7 10-2<\/sup> m
            \nT = 0.2 sec.<\/p>\n

            i) y = 5 cm = 5 \u00d7 10-2<\/sup> m
            \n\u03c9 = \\(\\frac{2 \\pi}{T}\\) = \\(\\frac{2 \\pi}{0.2}\\) = 10\u03c0
            \nAcceleration (a) = -\u03c92<\/sup>2y = -(10\u03c0)2<\/sup> \u00d7 5 \u00d7 10-2<\/sup>
            \na = -5\u03c02<\/sup> m\/s2<\/sup>
            \nVelocity (v) = \u03c9\\(\\sqrt{A^2-y^2}\\)
            \n= 10\u03c0\\(\\sqrt{\\left(5 \\times 10^{-2}\\right)^2-\\left(5 \\times 10^{-2}\\right)^2}\\)<\/p>\n

            ii) y = 3 cm = 3 \u00d7 10-2<\/sup> m
            \nAcceleration (a) = -\u03c92<\/sup>y = -(10\u03c0)2<\/sup> \u00d7 3 \u00d7 10-2<\/sup>
            \n= -3\u03c02<\/sup> m\/s2<\/sup>
            \nVelocity (v) = \u03c9\\(\\sqrt{A^2-y^2}\\)
            \n= 10\u03c0\\(\\sqrt{\\left(5 \\times 10^{-2}\\right)^2-\\left(3 \\times 10^{-2}\\right)^2}\\)
            \n= 10\u03c0\\(\\sqrt{25-9}\\) \u00d7 10-2<\/sup>
            \nv = 0.4\u03c0 m\/s.<\/p>\n

            iii) y = 0 cm
            \na = -\u03c92<\/sup>y = -(10\u03c0)2<\/sup> \u00d7 0
            \nVelocity (v) = \u03c9\\(\\sqrt{A^2-y^2}\\)
            \n= 10\u03c0\\(\\sqrt{\\left(5 \\times 10^{-2}\\right)^2-0}\\)
            \n= 0.5\u03c0 m\/s<\/p>\n

            Question 10.
            \nThe mass and radius of a planet are double that of the earth. If the time period of a simple pendulum on the earth is T, find the time period on the planet.
            \nSolution:
            \ng = \\(\\frac{\\mathrm{GM}}{\\mathrm{R}^2}\\)
            \ng \u221d \\(\\frac{M}{R^2}\\)
            \n\"AP<\/p>\n

            Question 11.
            \nCalculate the change in the length of a simple pendulum of length 1 m, when its period of oscillation changes from 2 s to 1.5 s. (T.S. Mar. ’18)
            \nSolution:
            \n\"AP<\/p>\n

            Question 12.
            \nA freely falling body takes 2 seconds to reach the ground on a plane, when it is dropped from a height of 8 m. If the period of a simple pendulum is seconds on the planet, calculate the length of the pendulum.
            \nSolution:
            \nu = 0, t = 2 sec, s = h = 8 m
            \ns = ut + \\(\\frac{1}{2}\\)at
            \ns = 0 \u00d7 t + \\(\\frac{1}{2}\\) \u00d7 g \u00d7 22<\/sup>
            \ng = 4m\/s2<\/sup>
            \nT = \\(2 \\pi \\sqrt{\\frac{1}{g}}\\)
            \n\u03c0 = \\(2 \\pi \\sqrt{\\frac{1}{4}}\\)
            \nl = 1 m = 100 cm.<\/p>\n

            \"AP<\/p>\n

            Question 13.
            \nThe period of a simple pendulum is found to increase by 50% when the length of the pendulum is increased by 0.6 m. Calculate the initial length and the initial period of oscillation at a place where g = 9.8 m s-2<\/sup>.
            \nSolution:
            \n\"AP<\/p>\n

            Question 14.
            \nA clock regulated by a second’s pendulum keeps correct time. During summer the length of the pendulum increases to 1.02 m. How much will the clock gain or lose in one day ?
            \nSolution:
            \nT = 2\u03c0\\(\\sqrt{\\frac{1}{g}}\\)
            \nT \u221d \\(\\sqrt{l}\\)
            \n\\(\\frac{d T}{T}\\) = \\(\\frac{1}{2} \\frac{\\mathrm{d} l}{l}\\)
            \nT = 2 sec, l = \\(\\frac{\\mathrm{g}}{\\pi^2}\\) = 0.9927
            \ndl = 1.02 – 0.9927 = 0.0273
            \n\\(\\frac{d T}{2}\\) = \\(\\frac{1}{2}\\left(\\frac{0.0273}{0.9927}\\right)\\)
            \ndT = \\(\\frac{0.0273}{0.9927}\\)
            \nNo. of oscillations performed per day by seconds pendulum = \\(\\frac{\\text { One day }}{2 \\text { Sec. }}\\)
            \n= \\(\\frac{86,400}{2}\\) = 43,200.<\/p>\n

            The gain (or) loss of time per day = No. of oscillations in one day to the change in time for one oscillation = 43,200 \u00d7 \\(\\frac{0.0273}{0.9927}\\)
            \n= 1180 sec.<\/p>\n

            Question 15.
            \nThe time period of a body suspended from a spring is T. What will be the new time period if the spring is cut into two equal parts and the mass is suspended
            \n(i) from one part
            \n(ii) simultaneously from both the parts ?
            \nAnswer:
            \nT = \\(2 \\pi \\sqrt{\\frac{m}{k}}\\)
            \ni) Spring is cut into two parts, k’ = 2k
            \nT’ = \\(2 \\pi \\sqrt{\\frac{m}{k^{\\prime}}}\\) = \\(\\frac{T}{\\sqrt{2}}\\)
            \nii) When the mass is suspended from both the parts
            \nT = \\(2 \\pi \\sqrt{\\frac{m}{2 k+2 k}}\\) = \\(2 \\pi \\sqrt{\\frac{m}{4 k}}\\) = \\(\\frac{T}{2}\\)<\/p>\n

            Additional Problems<\/span><\/p>\n

            Question 1.
            \nWhich of the following examples represent periodic motion ?
            \na) A swimmer completing one (return) trip from one bank of a river to the other and back. .
            \nb) A freely suspended bar magnet displaced from its N – S direction and released.
            \nc) A hydrogen molecule rotating about its centre of mass.
            \nd) An arrow released from a bow.
            \nSolution:
            \na) It is not a periodic motion. Though the motion of a swimmer is to and fro but will not have a definite period.
            \nb) It is a periodic motion because a freely suspended magnet if once displaced from N-S direction and let it go, it oscillation about this position. Hence it is simple harmonic motion also.
            \nc) It is also a periodic motion.
            \nd) It is not a periodic motion.<\/p>\n

            Question 2.
            \nWhich of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion ?
            \na) The rotation of earth about its axis.
            \nb) Motion of an oscillating mercury column in a U-tube.
            \nc) Motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower most point.
            \nd) General vibrations of a polyatomic molecule about its equilibrium position.
            \nSolution:
            \na) It is periodic but not SHM because it is not to and fro motion about a fixed point.
            \nb) It is SHM
            \nc) It is SHM
            \nd) It is a periodic but not SHM. A polyatomic gas molecule has a number of natural frequencies and its general motion is the resultant of S.H.M’s of a number of different frequencies. The resultant motion is periodic but not SHM.<\/p>\n

            \"AP<\/p>\n

            Question 3.
            \nFig. depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion ? What is the period of motion (in case of periodic motion)?
            \n\"AP
            \nSolution:
            \na) Does not represent periodic motion, as the motion neither repeats nor comes to mean position.
            \nb) Represents periodic motion with period equal to 2s.
            \nc) Does not represent periodic motion, because it is not identically repeated.
            \nd) Represents periodic motion with periodic equal to 2s.<\/p>\n

            Question 4.
            \nWhich of the following functions of time represent
            \n(a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion ? Give period for each case of periodic motion (\u03c9 is any positive constant)
            \na) sin \u03c9t – cos \u03c9t
            \nb) sin3<\/sup> \u03c9t
            \nc) 3 cos (\u03c0\/4 – 2\u03c9t)
            \nd) cos \u03c9t + cos 3\u03c9t + cos 5 \u03c9t
            \ne) exp (-\u03c92<\/sup>t2<\/sup>)
            \nf) 1 + \u03c9t + \u03c92<\/sup>t2<\/sup>.
            \nSolution:
            \nThe function will represent a periodic motion, if it is identically repeated after a fixed interval of time and will represent SHM. If it can be written uniquely in the form of a
            \ncos \\(\\left(\\frac{2 \\pi \\mathrm{t}}{\\mathrm{T}}+\\phi\\right)\\) or a
            \nsin \\(\\left(\\frac{2 \\pi \\mathrm{t}}{\\mathrm{T}}+\\phi\\right)\\), where T is the time period.<\/p>\n

            a) sin \u03c9t – cos \u03c9t
            \n= \\(\\sqrt{2}\\)(\\(\\frac{1}{\\sqrt{2}}\\)sin \u03c9t – \\(\\frac{1}{\\sqrt{2}}\\)cos \u03c9t)
            \n= \\(\\sqrt{2}\\)(sin \u03c9t cos \\(\\frac{\\pi}{4}\\) – cos \u03c9t sin \\(\\frac{\\pi}{4}\\))
            \n= \\(\\sqrt{2}\\)(\u03c9t\\(\\frac{\\pi}{4}\\))
            \nIt is a S.H.M and its period is 2\u03c0\/\u03c9.<\/p>\n

            b) sin3<\/sup> \u03c9t = \\(\\frac{1}{4}\\)(3sin \u03c9t – sin 3\u03c9t)
            \nHere each term sin \u03c9t and sin 3\u03c9t individually represents S.H.M. But (ii) which is the out come of the super position of two S.H.Ms will only be periodic but not S.H.M. its time period is 2\u03c0\/\u03c9.
            \n(\u2235 cos (-\u03b8) = cos \u03b8).<\/p>\n

            c) 3 cos \\(\\left(\\frac{\\pi}{4}-2 \\omega t\\right)\\) = 3 cos \\(\\left(2 \\omega t-\\frac{\\pi}{4}\\right)\\)
            \nClearly it represents S.H.M. and its time period is 2\u03c0\/2\u03c9.<\/p>\n

            d) cos \u03c9t + cos 3\u03c9t + cos 5\u03c9t it represents the periodic but not S.H.M its time period is 2\u03c0\/\u03c9.<\/p>\n

            e) \\(\\mathrm{e}^{-\\omega^2 t^2}\\) it is an exponential function which never repeats itself. Therefore it represents non-periodic motion.<\/p>\n

            f) 1 + \u03c9t + \u03c92<\/sup>t2<\/sup> also represents non periodic motion.<\/p>\n

            Question 5.
            \nA particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is
            \na) at the end A
            \nb) at end B
            \nc) at the mid-point of AB going towards A
            \nd) at 2 cm away from B going towards A
            \ne) at 3 cm away from A going towards B and
            \nf) at 4 cm away from B going towards A.
            \nSolution:
            \n\"AP
            \nRefer figure here A and B represent the two extreme positions of a S.H.M. For velocity, the direction from A to B is taken as positive for acceleration and the force, the direction is taken positive if directed along AP and negative if directed along BP.
            \na) At the end A, the artice executing S.H.M is momentarily at rest being its extreme position of motion, hence its velocity is zero, acceleration is +ve because directed along AP. Force is also +ve since the force is directed towards AP i.e + ve direction.
            \nb) At the end B, velocity is zero. Here acceleration and force are negative as they are directed along BP i.e. along negative direction.
            \nc) At the mid point AB going towards A, the particle is at its mean position P, with a tendency to move along PA i.e. -ve direction. Hence velocity is -ve both acceleration and force are zero.
            \nd) At 2 cm away from B going towards A, the particle is at Q, with a tendency to move along QP, which is a negative direction there velocity, acceleration and force are all -ve.
            \ne) At 3 cm away from A going towards B, the particle is at R with a tendency to move along RP, which is positive direction, there, velocity, acceleration and force are all +ve.
            \nf) At 4 cm away from A going towards A, the particle is at S, with a tendency to move along SA which is negative direction for velocity. Therefore velocity is negative but acceleration is directed towards mean position i.e., along SP, hence +ve. Similarly force is also +ve.<\/p>\n

            \"AP<\/p>\n

            Question 6.
            \nWhich of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion ?
            \na) a = 0.7x
            \nb) a = -200x2<\/sup>
            \nc) a = -10x
            \nd) a = 100x3<\/sup>
            \nSolution:
            \nIn S.H.M acceleration a is related to displacement by the related of the form a = -kx which is for relation (c).<\/p>\n

            Question 7.
            \nThe motion of a particle executing simple harmonic motion is described by the displacement function.
            \nx(t) = A cos (\u03c9t + \u03d5).
            \nIf the initial (t = 0) position of the particle is 1 cm and its initial velocity is \u03c9 cm\/s, what are its amplitude and initial phase angle ? The angular frequency of the particle is \u03c0s-1<\/sup>. If instead of the cosine function, we choose the sine function to describe the SHM : x = B sin (\u03c9t + \u03b1), what are the amplitude and initial phase of the particle with the above initial conditions.
            \nSolution:
            \nHere, at t = 0, x = 1 cm and v = \u03c9 cm s-1<\/sup>, \u03d5 = ?; \u03c9 = \u03c0s-1<\/sup>
            \nGiven x = A cos (\u03c9t + \u03d5)
            \n1 = A cos (\u03c0 \u00d7 0 + \u03d5) or
            \n= A cos \u03d5 —– (i)
            \nVelocity, v = \\(\\frac{d x}{d t}\\) = – A\u03c9(sin \u03c9t + \u03d5)
            \n\u2234 \u03c9 = -A\u03c9 sin (\u03c0 \u00d7 0 + \u03d5) or 1 = – A sin \u03d5
            \nor A sin \u03d5 = -1 —— (ii)
            \nSquaring and adding (i) and (ii)
            \nA2<\/sup>(cos2<\/sup> \u03d5 + sin2<\/sup> \u03d5) = 1 + 1 = 2 or A2<\/sup> = 2 or A = \\(\\sqrt{2}\\) cm
            \nDividing (ii) by (i), we get
            \ntan \u03d5 = -1 or \u03d5 = \\(\\frac{3 \\pi}{4}\\) or \\(\\frac{7 \\pi}{4}\\)
            \nFor, x = B sin (\u03c9t + \u03b1) —— (iii)
            \nAt t = 0, x = 1, so,
            \n1 = B sin (\u03c9 \u00d7 0 + \u03b1) = B sin \u03b1 —– (iv)
            \nDifferentiating (iii), w.r.t, t we have dx
            \nvelocity, v = \\(\\frac{d x}{d t}\\) = B\u03c9 cos (\u03c9t + \u03b1)
            \nApplying initial conditions i.e. at t = 0, v = \u03c9
            \n\u03c9 = B\u03c9cos (\u03c0 \u00d7 0 + \u03b1)
            \nor 1 = B cos \u03b1 —– (v)
            \nSquaring and adding (iv) and (v) we get
            \nB2<\/sup>sin2<\/sup>\u03b1 + B2<\/sup> cos2<\/sup> \u03b1 = 12<\/sup> + 12<\/sup> = 2 or B2<\/sup> = 2 or B = \\(\\sqrt{2}\\) cm
            \nDividing (iv) by (v), we have
            \n\\(\\frac{B \\sin \\alpha}{B \\cos \\alpha}\\) = \\(\\frac{1}{1}\\) or tan \u03b1 = 1 or \u03b1 = \\(\\frac{\\pi}{4}\\) or \\(\\frac{5 \\pi}{4}\\).<\/p>\n

            Question 8.
            \nA spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body ?
            \nSolution:
            \nHere. m = 50 kg, max. extension.
            \ny = 20 – 0 = 20 cm = 0.2 m; T = 0.65
            \nMax. Force, F = mg = 5 \u00d7 9.8 N
            \nK = \\(\\frac{F}{y}\\) = \\(\\frac{50 \\times 9.8}{0.2}\\) = 2450 Nm-1<\/sup>
            \nAs T = \\(2 \\pi \\sqrt{\\frac{m}{k}}\\)
            \nm = \\(\\frac{T^2 k}{4 \\pi^2}\\)
            \n= \\(\\frac{(0.6)^2 \\times 2450}{4 \\times(3.14)^2}\\)
            \n= 22.36 kg
            \n\u2234 Weight of body = mg = 22.36 \u00d7 9.8
            \n= 219.1 N
            \n= 22.36 kgf<\/p>\n

            \"AP<\/p>\n

            Question 9.
            \nA spring having with a spring constant 1200 Nm-1<\/sup> is mounted on a horizontal table as shown in fig. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.
            \n\"AP
            \nDetermine
            \n(i) the frequency of oscillations,
            \n(ii) maximum acceleration of the mass, and
            \n(iii) the maximum speed of the mass.
            \nSolution:
            \nHere, k = 1200 Nm-1<\/sup>, m = 3.0 kg, a = 2.0 cm = 0.02 m
            \na)
            \nFrequency,
            \n\u03c5 = \\(\\frac{1}{T}\\) = \\(\\frac{1}{2 \\pi} \\sqrt{\\frac{k}{m}}\\)
            \n= \\(\\frac{1}{2 \\times 3.14} \\sqrt{\\frac{1200}{3}}\\)
            \n= 3.2s-1<\/sup><\/p>\n

            b) Acceleration A = \u03c92<\/sup>y = \\(\\frac{k}{m}\\)y
            \nAcceleration will be maximum when y is maximum i.e. y = a
            \nMax. acceleration, Amax<\/sub> = \\(\\frac{k_a}{m}\\) = \\(\\frac{1200 \\times 0.02}{3}\\)
            \n= 8 ms-2<\/sup><\/p>\n

            c) Max. speed of the mass will be when it is passing through the mean position, which is given by
            \nVmax<\/sub> = a\u03c9 = a\\(\\sqrt{\\frac{k}{m}}\\)
            \n= 0.02 \u00d7 \\(\\sqrt{\\frac{1200}{3}}\\) = 0.4 ms-1<\/sup><\/p>\n

            Question 10.
            \nIn exercise, let us take the position of mass when the spring is unstreched as x = 0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is
            \na) at the mean position
            \nb) at the maximum stretched position and
            \nc) at the maximum compressed position
            \nIn what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase ?
            \nSolution:
            \nHere, a = 2.0 cm; \u03c9 = \\(\\sqrt{\\frac{k}{m}}\\) = \\(\\sqrt{\\frac{1200}{3}}\\) = 20s-1<\/sup><\/p>\n

            a) As time is noted from the mean position, hence using
            \nx = a sin \u03c9t, we have x = 2 sin 20t<\/p>\n

            b) At maximum stretched position, the body is the extreme right position, with an initial phase of \\(\\frac{\\pi}{2}\\) rad. Then
            \nx = a sin \\(\\left(\\omega t+\\frac{\\pi}{2}\\right)\\)
            \n= a cos \u03c9t = 2 cos 20t<\/p>\n

            c) At maximum compressed position, the body is at the extreme left position, with an initial phase of \\(\\frac{3 \\pi}{2}\\) rad.
            \nThen x = a sin (\u03c9t + \\(\\frac{3 \\pi}{2}\\))
            \n= -a cos \u03c9t
            \n= -2 cos 20t<\/p>\n

            Question 11.
            \nFigures correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anti-clockwise) are indicated on each figure.
            \n\"AP
            \nObtain the corresponding simple ‘ harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.
            \nSolution:
            \nIn Fig. (a) T = 2s ; a = 3 cm
            \nAt t = 0, OP makes an angle \\(\\frac{\\pi}{2}\\) with X-axis
            \ni.e., \u03d5 = \\(\\frac{\\pi}{2}\\) radian.
            \nWhile moving clockwise, here \u03d5 = +\\(\\frac{\\pi}{2}\\). Thus the X-projection of OP at time t will give us the equation of S.H.M. given by
            \nx = a cos \\(\\left(\\frac{2 \\pi t}{T}+\\phi\\right)\\)
            \n= 3cos \\(\\left(\\frac{2 \\pi t}{2}+\\frac{\\pi}{2}\\right)\\) or x = -3
            \nsin \u03c0t (x is in cm)
            \nIn Fig. (b) T = 4s ; a = 2m
            \nAt t = 0, OP makes an angle \u03c0 with the positive direction of X-axis i.e., \u03d5 = \u03c0. While moving anticlockwise, here \u03d5 = +\u03c0.
            \nThus the X-projection of OP at time t will give us the equation of S.H.M. as
            \nx = a cos \\(\\left(\\frac{2 \\pi t}{T}+\\phi\\right)\\)
            \n= 2 cos \\(\\left(\\frac{2 \\pi t}{4}+\\pi\\right)\\)
            \n= -2 cos \\(\\left(\\frac{\\pi}{2} t\\right)\\) (x is in m)<\/p>\n

            Question 12.
            \nPlot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t = 0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case : (x is in cm and t is in s).
            \na) x = -2 sin (3t + \u03c0\/3)
            \nb) x = cos (\u03c0\/6 – t)
            \nc) x = 3 sin (2\u03c0t + \u03c0\/4)
            \nd) x = 2 cos \u03c0t.
            \nSolution:
            \nIf we express each function of the form
            \nx = a cos (cot + \u03d5) —— (i)
            \nWhere \u03d5 is the initial phase i.e., \u03d5 represents the angle which the initial radius vector of the particle makes with the -l-ve direction of X-axis.<\/p>\n

            a) x = -2 sin (3t + \\(\\frac{\\pi}{3}\\)) = 2 cos (\\(\\frac{\\pi}{2}\\) + 3t + \\(\\frac{\\pi}{3}\\))
            \nor x = 2 cos (3t + \\(\\frac{5 \\pi}{6}\\))
            \nComparing it with equation (i), we note
            \nthat a = 2, \u03c9 = 3 and \u03d5 = \\(\\frac{5 \\pi}{6}\\)
            \nHence, the reference circle will be shown in Fig. (a).<\/p>\n

            b) x = cos\\(\\left(\\frac{\\pi}{6}-t\\right)\\) = cos \\(\\left(t-\\frac{\\pi}{6}\\right)\\)
            \n(\u2235 cos (-\u03b8) = cos \u03b8))
            \nComparing it with equation (i), we note
            \nthat a = 1, \u03c9 and \u03d5 = \\(\\frac{\\pi}{6}\\)
            \nThe reference circle will be as shown in Fig. (b).<\/p>\n

            c) x = 3 sin \\(\\left(2 \\pi t+\\frac{\\pi}{4}\\right)\\)
            \n= 3 cos \\(\\left(2 \\pi t+\\frac{3 \\pi}{2}+\\frac{\\pi}{4}\\right)\\)
            \nComparing it with equation (i), we note that a = 3, \u03c9 = 2\u03c0 and \u03d5 = \\(\\frac{3 \\pi}{2}\\) + \\(\\frac{\\pi}{4}\\) = \\(\\frac{7 \\pi}{4}\\)
            \nThe reference circle will be as in Fig. (c).<\/p>\n

            d) x = 2 cos \u03c0t
            \nComparing it with equation (i), we note
            \nthat a = 2, = \u03c0 and \u03d5 = 0.
            \nThe reference circle will be as shown Fig. (d)
            \n\"AP<\/p>\n

            \"AP<\/p>\n

            Question 13.
            \nFigures (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure (b) shows the same spring with both ends free and attached to a mass at either end. Each end of the spring in Fig. (b) is stretched by the same force F.
            \n\"AP
            \na) What is the maximum extension of the spring in two cases ?
            \nb) If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case ?
            \nSolution:
            \n\"AP
            \na) The maximum extension of the spring in both the cases will be = \\(\\frac{F}{K}\\), where K is the spring constant of the spring used.
            \nb) In Fig. (a), if x is the extension in the spring, when mass m is returning to its mean posi-tion after being released free, then restoring force on the mass is F = -Kx i.e., F \u221d x.
            \nAs this F is directed towards mean position of the mass, hence the mass attached to the spring will execute SHM. Here, spring factor = spring constant
            \n= K
            \nInertia factor = mass of the given mass
            \n= m
            \n\"AP
            \nIn Fig.(b). we have a two body system of spring constant K and reduced mass,
            \n\u03bc = \\(\\frac{m \\times m}{m+m}\\) = \\(\\frac{m}{2}\\)
            \nHere, inertia factor = \\(\\frac{m}{2}\\)
            \nand spring factor = K
            \n\u2234 Time period, T = 2\u03c0\\(\\sqrt{\\frac{(m \/ 2)}{k}}\\)
            \n= 2\u03c0\\(\\sqrt{\\frac{m}{2 K}}\\)<\/p>\n

            Question 14.
            \nThe piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad\/min. what is its maximum speed ?
            \nSolution:
            \nGiven a = \\(\\frac{1}{2}\\)m; \u03c9 = 200 rev\/min ;
            \nVmax<\/sub> = a\u03c9
            \n= \\(\\frac{1}{2}\\) \u00d7 200
            \n= 100 m\/min.<\/p>\n

            Question 15.
            \nThe acceleration due to gravity on the surface of moon is 1.7 ms-2<\/sup>. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s ? (g on the surface of earth is 9.8 ms-2<\/sup>).
            \nSolution:
            \n\"AP<\/p>\n

            Question 16.
            \nAnswer the following questions :
            \na) Time period of a particle in SHM depends on the force constant k and mass m of the particle :
            \nT = \\(2 \\pi \\sqrt{\\frac{\\mathrm{m}}{\\mathrm{k}}}\\). A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum ?
            \nb) The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation, a more involved analysis shows that T is greater than \\(2 \\pi \\sqrt{\\frac{l}{g}}\\). Think of a qualitative argument to appreciate this result.
            \nc) A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall ?
            \nd) What is the frequency of oscillation of a simple pendulum mounted in cabin that is freely falling under gravity ?
            \nSolution:
            \na) For a simple pendulum, force constant or spring factor K is proportional to mass m, therefore, m cancels out in denominator as well as in numerator. That is why the time period of simple pendulum is independent of the mass of the bob.<\/p>\n

            b) The effective restoring force acting on the bob of simple pendulum in displaced position is
            \nF = -mg sin \u03b8. When \u03b8 is small, sin \u03b8 = \u03b8. Then the expression for time period of simple pendulum is given by T = \\(2 \\pi \\sqrt{\\mu \\mathrm{g}}\\) When \u03b8 is large sin \u03b8 < \u03b8, if the restoring force mg sin \u03b8 is replaced by mg\u03b8, this amounts to effective reduction in the value of ‘g’ for large angles and hence an increase in the value of time period T.<\/p>\n

            c) Yes, because the working of the wrist watch depends on spring action and it has nothing to do with gravity.<\/p>\n

            d) We know that gravity disappears for a man under free fall, so frequency is zero.<\/p>\n

            Question 17.
            \nA simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period ?
            \nSolution:
            \nCentripetal acceleration, ac<\/sub> = \\(\\frac{v^2}{R}\\), it is acting horizontally.
            \nAcceleration due to gravity = g acting vertically downwards.
            \nEffective acceleration due to gravity
            \ng’ = \\(\\sqrt{g^2+\\frac{v^4}{R^2}}\\)
            \n\u2234 Time period, T = 2\u03c0\\(\\sqrt{\\frac{1}{g}}\\)
            \n= 2\u03c0\\(\\sqrt{\\frac{1}{g^2+v^4 \/ R^2}}\\)<\/p>\n

            \"AP<\/p>\n

            Question 18.
            \nA cylindrical piece of cork of density of base area A and height h floats in a liquid of density \u03c11<\/sub>. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period.
            \nT = \\(2 \\pi \\sqrt{\\frac{h \\rho}{\\rho_1 g}}\\)
            \nwhere \u03c1 is the density of cork. (Ignore damping due to viscosity of the liquid).
            \nSolution:
            \nMass of the cylinder (m) = volume \u00d7 density
            \n= Ah\u03c1 ——- (1)
            \nF1<\/sub> = weight of the liquid displaced by the length l of the cylinder
            \n= (Al)\u03c11<\/sub>g —— (2)
            \nWeight of the cylinder = mg —– (3)
            \nIn equilibrium position, mg = Al\u03c11<\/sub>g
            \nm = Al\u03c11<\/sub> —– (4)
            \nF2<\/sub> = A(l + y)\u03c11<\/sub>g —– (5)
            \n\"AP
            \nRestoring force (F) = -(F2<\/sub> – mg)
            \n= -[A(l + y)\u03c11<\/sub>g – Al\u03c11<\/sub>g]
            \nF = Ay\u03c11<\/sub>g = -(A\u03c11<\/sub>g)y —— (6)
            \nIn S.H.M. F = -Ky —– (7)
            \nFrom eqs. (6) & (7),
            \nspring factor (K) = A\u03c11<\/sub>g ——(8)
            \nInertia factor, m = Ah\u03c1 ——- (9)
            \n\"AP<\/p>\n

            Question 19.
            \nOne end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.
            \nSolution:
            \nConsider a liquid of density \u03c1 contained in a vertical U-tube of cross-sectional area A. Total length of the liquid column from P to P1<\/sub> is L.
            \n\"AP
            \nMass (m) = LA\u03c1
            \nPQ = y, P1<\/sub>Q1<\/sub> = y, QQ1<\/sub> = 2y
            \nRestoring force (F) = -(A2y)\u03c1g
            \n= -(2A\u03c1g)y —– (1)
            \nF \u221d -y
            \nHence oscillations in U-tube is S.H.M.<\/p>\n

            Question 20.
            \nAn air chamber of volume V has a neck area of cross section a into which a ball of mass m just fits and can move up and down without any friction (Fig.). Show that when the ball is pressed down a little and released, it executes SHM. Obtain an expression for the time period of oscillations assuming pressure-volume variations of air to be isothermal (see Fig.).
            \n\"AP
            \nSolution:
            \nConsider an air chamber of volume V with a long neck of uniform area of cross-section A, and a frictionless ball of mass m filled smoothly in the neck at position C. The pressure of air below the ball inside the chamber is equal to the atmospheric pressure. Increase the pressure on the ball by a little amount P, so that the ball is depressed to position D, where CD = y.
            \nThere will be decrease in volume and hence increases in pressure of air inside the chamber. The decrease in volume of the air inside the chamber.
            \n\"AP
            \nHere, negative sign shows that the increase in pressure will decrease the volume of air in the chamber.
            \nNow, \u03c1 = \\(\\frac{-E A y}{V}\\)
            \nDue to this excess pressure, the restoring force acting on the ball is
            \nF = P \u00d7 A = \\(\\frac{-E A y}{V} \\cdot A\\)
            \n= \\(\\frac{-E A^2}{V} y\\) —- (1)
            \nSince F \u221d y and negative sign shows that the force is directed towards equilibrium position, if the applied increased pressure is removed from the ball, the ball with start executing linear SHM in the neck of chamber with C as mean position.
            \nIn a S.H.M., the restoring force,
            \nF = -Ky —- (2)
            \nComparing (1) and (2), we have spring factor.
            \nK = \\(\\frac{E A^2}{V}\\)
            \nHere, inertia factor = mass of ball = m
            \nInertia factor Spring factor
            \n\"AP<\/p>\n

            Note : If the ball oscillates in the neck of chamber under isothermal conditions, then E = P = Pressure of air inside the chamber, when ball is at equilibrium position. If the ball oscillates in the neck of chamber under adiabatic conditions, then E = \u03c5P, where v = \\(\\frac{c_p}{c_v}\\).<\/p>\n

            Question 21.
            \nYou are riding in an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of (a) the spring constant k and (b) the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg.
            \nSolution:
            \nHere, M = 3000 kg ; x = 0.15 cm ; if K is the spring constant of each spring, then spring constant of 4 springs in parallel to support the whole mass is, K = 4 K.
            \n4 kx = Mg
            \nk = \\(\\frac{M g}{4 x}\\)
            \n= \\(\\frac{3000 \\times 10}{4 \\times 0.15}\\)
            \n= 5 \u00d7 104<\/sup> N\/m.<\/p>\n

            b) If m is the mass supported by each spring, then m = \\(\\frac{3000}{4}\\) = 750 kg.
            \n\"AP<\/p>\n

            Question 22.
            \nShow that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.
            \nSolution:
            \nConsider a particle of mass m be executing S.H.M. with period T. The displacement of the particle at an instant t, when time period is noted from the mean position is given by
            \nY = a sin \u03c9t
            \n\u2234 Velocity 1 V = \\(\\frac{\\mathrm{dy}}{\\mathrm{d} \\mathrm{t}}\\) = a\u03c9 cos \u03c9t
            \nK.E., EK<\/sub> = \\(\\frac{1}{2}\\)mv2<\/sup> = \\(\\frac{1}{2}\\)ma2<\/sup>\u03c92<\/sup> cos2<\/sup> \u03c9t
            \n\"AP
            \nAverage P.E. over one cycle is
            \n\"AP<\/p>\n

            \"AP<\/p>\n

            Question 23.
            \nA circular disc of mass 10 kg is suspended by a wire attached to is centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant a is defined by the relation J = -\u03b1 \u03b8, where J is the restoring couple and \u03b8 the angle of twist).
            \nSolution:
            \nHere, m = 10kg; R = 15 cm = 0.15m; T = 1.55, \u221d = ?
            \nMoment of inertia of disc,
            \nI = \\(\\frac{1}{2} m R^2\\)
            \n= \\(\\frac{1}{2}\\) \u00d7 10 \u00d7 (0.15)2<\/sup> kgm2<\/sup>
            \nNow T = \\(2 \\pi \\sqrt{\\frac{1}{\\alpha}}\\)
            \nso, \u03b1 = \\(\\frac{4 \\pi^2 1}{T^2}\\)
            \n= 4 \u00d7 \\(\\left(\\frac{22}{7}\\right)^2\\) \u00d7 \\(\\frac{1}{2}\\) \u00d7 \\(\\frac{10 \\times(0.15)^2}{(1.5)^2}\\)
            \n= 1.97Nm\/rad.<\/p>\n

            Question 24.
            \nA body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is
            \n(a) 5 cm
            \n(b) 3 cm
            \n(c) 0 cm.
            \nSolution:
            \nHere, r = 5 cm = 0.05 m ; T = 0.25 ;
            \n\u03c9 = \\(\\frac{2 \\pi}{\\mathrm{T}}\\) = \\(\\frac{2 \\pi}{0.2}\\)
            \n= 10p rad\/s
            \nWhen displacement is y, then acceleration
            \nA = -\u03c92<\/sup>y
            \nVelocity, v = \u03c9\\(\\sqrt{r^2-y^2}\\)<\/p>\n

            Case (a) : When y = x cm = 0.05 m
            \nA = -(10\u03c0)2<\/sup> \u00d7 0.05
            \n= -5\u03c02<\/sup> m\/s2<\/sup>
            \nV = 10\u03c0\\(\\sqrt{(0.05)^2-(0.05)^2}\\) = 0<\/p>\n

            Case (b) : When y = 3 cm = 0.03 m
            \nA = -(10\u03c0)2<\/sup> \u00d7 0.03
            \n= -3\u03c02<\/sup> m\/s2<\/sup>
            \nV= 10\u03c0 \u00d7 \\(\\sqrt{(0.05)^2-(0.03)^2}\\)
            \n= 10\u03c0 \u00d7 0.04
            \n= 0.4\u03c0 m\/s<\/p>\n

            Case (c): When y = 0,
            \nA = -(10\u03c0)2<\/sup> \u00d7 0 = 0
            \nV= 10\u03c0\\(\\sqrt{(0.05)^2-(0)^2}\\)
            \n= 10\u03c0 \u00d7 0.05
            \n= 0.5\u03c0 m\/s.<\/p>\n

            Question 25.
            \nA mass attached to a spring is free to oscillate, with angular velocity \u03c9, in a horizontal plane without friction or damping. It is pulled to a distance x0<\/sub> and pushed towards the centre with a velocity v0<\/sub> at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters \u03c9, x0<\/sub> and V0<\/sub>.
            \n[Hint : Start with the equation x = a cos (\u03c9t + \u03b8) and note that the initial velocity is negative.]
            \nSolution:
            \nx = A cos (\u03c9t + \u03b8) dx
            \nVelocity, \\(\\frac{\\mathrm{dx}}{\\mathrm{dt}}\\) = -A\u03c9 sin (\u03c9t + \u03b8) dt
            \nWhen t = 0, x = x0<\/sub>, and \\(\\frac{\\mathrm{dx}}{\\mathrm{dt}}\\) = -V0<\/sub>
            \n\u2234 x0<\/sub> = A cos \u03b8
            \n-V0<\/sub> = -A\u03c9 sin \u03b8 or A sin \u03b8 = \\(\\frac{V_0}{\\omega}\\)
            \nSquaring and adding- (i) and (ii), we get
            \nA2<\/sup>(sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8) = \\(\\left(\\frac{v_0^2}{\\omega^2}\\right)+x_0^2\\)
            \nA = \\(\\left[\\frac{v_0^2}{\\omega^2}+x_0^2\\right]^{1 \/ 2}\\)<\/p>\n","protected":false},"excerpt":{"rendered":"

            Andhra Pradesh BIEAP\u00a0AP Inter 1st Year Physics Study Material 8th Lesson Oscillations Textbook Questions and Answers. AP Inter 1st Year Physics Study Material 8th Lesson Oscillations Very Short Answer Questions Question 1. Give two examples of periodic motion which are not oscillatory. Answer: The motion of planets around the sun. The motion of an electron … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":39424,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[13],"tags":[],"jetpack_featured_media_url":"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/09\/Chapter-8-Oscillations-1.png","_links":{"self":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/28717"}],"collection":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/comments?post=28717"}],"version-history":[{"count":18,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/28717\/revisions"}],"predecessor-version":[{"id":28861,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/28717\/revisions\/28861"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media\/39424"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media?parent=28717"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/categories?post=28717"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/tags?post=28717"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}