Oscillations Questions and Answers AP Inter 1st Year Physics Chapter 13

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 13th Lesson Oscillations Class 11 Textbook Exercise Questions and Answers.

Oscillations Class 11 Questions and Answers AP Inter 1st Year Physics 13th Lesson

I. Oscillations Multiple Choice Questions (1 Mark)

Question 1.
Which one of the following represents the motion that is periodic but not simple harmonic?
(1) Motion of the earth around the sun
(2) Oscillations of a loaded spring
(3) Oscillations of a balance wheel of watch
(4) Oscillations of liquid level in U-tube
Answer:
(1) Motion of the earth around the sun

Question 2.
Identify the periodic function in the following.
(1) e^ωt
(2) e^-ωt
(3) log (ωt)
(4) sin ωt+cos ωt
Answer:
(4) sin ωt+cos ωt

Question 3.
The unit of frequency is……………….
(1) sec
(2) rad/sec
(3) m/s
(4) sec^-1
Answer:
(4) sec^-1

Question 4.
If the displacement of the particle in SHM is Y(t) = Asinωt+B cos ωt, then the amplitude of its motion is
(1) A²+B²
(2) A+B
(3) \(\sqrt{\left(\mathrm{A}^2+\mathrm{B}^2\right)}\)
(4) AB
Answer:
(3) \(\sqrt{\left(\mathrm{A}^2+\mathrm{B}^2\right)}\)

Question 5.
The quantity which doesn’t vary periodically in SHM is…………..
(1) displacement
(2) velocity
(3) acceleration
(4) total energy
Answer:
(4) total energy

Question 6.
The expression for the maximum velocity in SHM is…………
(1) ωA
(2) ω²A
(3) ωA²
(4) ω²A²
Answer:
(2) ω²A

Question 7.
The total energy of a Simple harmonic oscillator is…………
(1) \(\frac{1}{2} \mathrm{KA}^2\)
(2) \( \frac{1}{2} K^2 A\)
(3) \( \frac{1}{2} \mathrm{~K}^2 \mathrm{~A}^2\)
(4) \(\frac{1}{2} \mathrm{KA}\)
Answer:
(1) \(\frac{1}{2} \mathrm{KA}^2\)

Question 8.
The total energy of a simple harmonic oscillator is proportional to………….
(1) square of the Frequency
(2) velocity
(3) amplitude
(4) displacement
Answer:
(1) square of the Frequency

Question 9.
The magnitude of maximum acceleration of a simple harmonic oscillator is given by………..
(1) ωA
(2) ω²A
(3) ωA²
(4) ω²A²
Answer:
(2) ω²A

Question 10.
The time period of a simple pendulum is proportional to
(1) length of the pendulum
(2) mass of the pendulum
(3) velocity of the pendulum
(4) square root of the length of the pendulum
Answer:
(4) square root of the length of the pendulum

II. Oscillations Fill in the Blanks (1 Mark)

Question 1.
For a simple harmonic oscillator, the restoring force is directly proportional to …………..
Answer:
Displacement

Question 2.
The general equation for simple harmonic motion is …………..
Y=A sin (ωt+φ), or X=A cos (ωt+φ)

Question 3.
In SHM, the value of acceleration at the mean position is …………..
Answer:
Zero

Question 4.
The potential energy of a Simple harmonic oscillator is maximum at ………….. position.
Answer:
Extreme position

Question 5.
The kinetic energy of Simple harmonic oscillator is maximum at ………….. position.
Answer:
Mean position

Question 6.
The equation for the time period of a simple pendulum is …………..
Answer:
\(T=2 \pi \sqrt{1/g}\)

Question 7.
If the length of the pendulum is 100 cm, then that pendulum is called …………..
Answer:
Seconds pendulum

Question 8.
The time period of a seconds pendulum is …………..
Answer:
Two seconds

Question 9.
The frequency of a seconds pendulum is …………..
Answer:
0.5 Hz

III. Oscillations One Word Answer Questions (1 Mark)

Question 1.
What is a periodic motion?
Answer:
A motion that repeats itself ofter equal intervals of time is known as periodic motion.

Question 2.
Give any one example of a body whose motion is periodic but not oscillatory.
Answer:
The Earth’s orbit around the sun.

Question 3.
What is the reciprocal of the time period?
Answer:
Frequency

Question 4.
Define amplitude.
Answer:
Loudness

Question 5.
Express the relation between angular frequency (ω) and time period (T).
Answer:
ω = \(\frac{2 \pi}{T}\)

Question 6.
What is the phase difference between displacement and acceleration of a particle executing simple harmonic motion?
Answer:
Opposing

Question 7.
What is the phase difference between displacement and velocity of a particle executing simple harmonic motion?
Answer:
π/2 (or) 90°

Question 8.
What is the phase difference between velocity and acceleration of a particle executing simple harmonic motion?
Answer:
90° (or) π/2

Question 9.
What is the maximum Kinetic energy of a simple harmonic oscillator ?
Answer:
The maximum kinetic energy is equal to the total energy of the system.

Question 10.
What is the value of time period of a seconds pendulum?
Answer:
Two seconds

Question 11.
What is the time period of a simple pendulum in an artificial satellite ?
Answer:
Infinite

IV. Oscillations Very Short Answer Questions (2 Marks)

Question 1.
Give two examples of periodic motion which are not oscillatory.
Answer:

• Motion of seconds hand of a watch.
• Revolution of a planet around the sun.
• Revolution of an electron around the nucleus.
• Revolution of the moon around the earth.

Question 2.
The displacement in SHM is given by y=a sin (20t + 4). What is the displacement when it is increased by 2 π/ω?
Answer:
The given equation y=a sin (20 t+4) is similar to y=a sin (ωt+4).
If t is increased by \(\frac{2 \pi}{\omega}\) or \(\frac{2 \pi}{20}\) then \(\frac{2 \pi}{\omega}\) = T, with in time period T, the particle regains to the same position. Hence, the displacement does not change.

Question 3.
A girl is swinging seated in a swing. What is the effect on the frequency of oscillation, if she stands?
Answer:
From time period T =2 π \(\sqrt{\frac{l}{g}}\) and at same place Tα √l and frequency nα  \(\frac{1}{\sqrt{l}}\)
If a girl sitting in a swing stands up, then centre of mass shifts up and hence length ‘l’ of the swing decreases. Then the time period decreases and frequency of oscillations increases.

Question 4.
The bob of a simple pendulum is a hallow sphere filled with water. How will the period of oscillation change, if the water begins to drain out of the hallow sphere?
Answer:
From time period T=2π \(\sqrt{\frac{l}{g}}\) and at same place T α √l. If water begins to drainout from the hollow sphere, then centre of mass of (sphere + water) the system goes down and hence effective length ‘ l ‘ of the pendulum increases.

Then time period increases. If water is completely drained out from the sphere then the centre of mass comes to its orginal position (i.e., centre of the sphere). So the pendulum will have its original time period.

Question 5.
The 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 aluminium?
Answer:
From time period T=2 π \(\sqrt{\frac{l}{g}}\) and at same place T α √l.

If the wooden bob is replaced by an identical bob of aluminum then ‘ l ‘ remains constant, so time period ‘ T ‘ does not change. Here mass and material of the bob will not change time period.

Question 6.
Will a pendulum clock gain or lose time when taken to the top of a mountain?
Answer:
From time period T= \(2 \pi \sqrt{\frac{l}{g}}\) and for the same pendulum Tα \(\frac{1}{\sqrt{g}}\) and n=\(\frac{1}{\mathrm{~T}}\)
When the pendulum is taken to the top of a mountain, acceleration due to gravity ‘g’ decreases, time period increases and number of oscillations per second decreases. So the number of oscillations per day also decreases. Hence it looses time.

Question 7.
A pendulum clock gives correct time at the equator. Will it gain or lose time if it is taken to the poles? If so, why?
Answer:
From time period T = \(2 \pi \sqrt{\frac{l}{g}}\) and for the same pendulum T α \(\frac{1}{\sqrt{g}}\) and n = \(\frac{1}{\mathrm{~T}}\)
When a pendulum is taken to the poles from equator then acceleration due to gravity increases, time period decreases and number of oscillations per second increases. So the number of oscillations per a day also increases. Hence the clock gains time.

Question 8.
What fraction of the total energy is Kr: when the displacement is half of the amplitude of a particle executing S.H.M. ?
Answer:
We know that,

Question 9.
What happens to the energy of a simple harmonic oscillator if its amplitude is doubled?
Answer:
Energy of a simple harmonic oscillator E= \(\frac{1}{2}\) mω²A²
For the same oscillation ‘m’ and ‘ω’ are constants then

∴ The energy becomes 4 times.

Question 10.
Can a simple pendulum be used in an artificial satellite ? Why?
Answer:
No.
In an artificial satellite

So, pendulum doesn’t oscillate.

V. Oscillations Short Answer Questions (4 Marks)

Question 1.
Define simple harmonic motion ? Give two examples.
Answer:
Definition:
A body is said to be in simple harmonic motion provided.

• The motion should be ‘to and fro’ and is periodic about a fixed point.
• The acceleration should always be directed towards the mean position and is always directly proportional to the displacement from the mean position.
• Acceleration and displacement are always in opposite direction.

Example :

• Oscillations of a loaded spring suspended from a rigid support.
  K =spring constant.
• The projection of a particle performing uniform circular motion on any diameter.
• Oscillation of a liquid column in a U tube.

Question 2.
Present graphically the variations of displacement, velocity and acceleration with time for a particle in SHM.
Answer:

X varies between -A to A.
v(t) varies from -ωA to ωA
a(t) varies from -ω²A to ω²A with respect to displacement plot,

• Velocity plot has a phase difference of π/ 2 and
• Acceleration plot has a phase difference of π.

Question 3.
What is phase? Discuss the phase relations between displacement, velocity and acceleration in simple harmonic motion.
Answer:
Phase:
The phase of a particle executing SHM discribes its postition and direction of motion at any instant. Phase relationship among the displacement (y), velocity (v) and acceleration (a) in S.H.M.
When the particle starts from the mean position y=A sinωt ……….. (1)
Its velocity, v = \(\frac{\mathrm{dy}}{\mathrm{dt}}\)=Aωcos (ωt)
Its acceleration, a =\(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dt}^2}\)=-Aω²sin (ωt)
∴ a=- ω²y

• v leads ‘y’ by π/ 2
• a lags behind ‘ v ‘ by π/2
• ‘a’ differes in phase with ‘y’ by π

Question 4.
How does the energy of a simple pendulum vary as it moves from one extreme position to the other during its oscillations?
Answer:
At the extreme positions of the simple pendulum, the velocity of the bob is zero. Hence its K.E is zero, but PE is maximum.

Hence total energy is P.E
At the mean position, the velocity of the bob is maximum and so its K.E is maximum. But PE is zero. Hence total energy is only K.E.

At any point between mean and extreme positions, the total energy is the sum of P.E and K.E. So total energy of the pendulum at any position is constant.

Both KE and PE in SHM are always +ve.

Both PE and KE get peak values twice during each period of SHM.
For x=0, the energy is KE,
At extremes x= ± A, the energy is PE.
KE increases at the expense of potential energy and vice-versa.

Question 5.
Derive the expression for displacement, velocity and acceleration of a particle executes S.H.M.
Answer:
Consider a particle P moving on a circle of radius A with uniform angular speed ω in anti clock wise direction. As the particle moves on the circumference of the circle, its projection M moves on the diameter xx’ to and fro about centre ‘0’. Let the particle starts from the position ‘x’ After time ‘t’ the position vector of the particle covers an angle of ‘ωt ‘ with +ve x -axis. During this time the projection (M) of position vector of the particle on x-axis covers a displacement x.

Displacement :
It is the distance travelled by foot of the perpendicular in time ‘t’
From the triangle OMP, cos ωt = \(\frac{\mathrm{x}}{\mathrm{A}}\)
The displacement x=A cos ωt ………….(1)

Velocity:
The rate of change of displacement is called velocity.

Acceleration :
The rate of change of velocity of a particle is called acceleration.

VI. Oscillations Long Type Questions (8 Marks)

Question 1.
Derive an expression for the Kinetic energy and potential energy of a simple harmonic oscillator.
Answer:
Any body or system executing SHM is called simple harmonic oscillator.
Expression for kinetic Energy :
Consider a particle of mass m executing SHM between extreme positions x=+A and x=- A. Displacement of the particle in SHM when it starts from extreme position is given by
x=A cos ωt
Velocity of the particle executing SHM is given by

Expression for Potential Energy :
When a particle executing SHM is displaced through ‘x’ from equilibrium position, then restoring force acting on it is F=-kx. Further, if it displaces through ‘dx’ then workdone against the restoring force is given by
dW =-Fdx
⇒ dW =-(-kxdx)
∴ dW=(kxdx)
∴ Total workdone against the restoring force in moving the particle from equilibrium position (x=0) to any displacement (x) is given by

Question 2.
Define simple harmonic motion. Show that the motion of (point) projection of a particle performing uniform circular motion, on any diameter, is simple harmonic.
Answer:
“A body is said to be in simple harmonic motion provided.

• The motion should be to and fro and its periodic about a fixed point.
• The acceleration should always be directed towards the mean position and is always directly proportional to the displacement from the mean position.
• Acceleration and displacement are always in opposite direction.

Consider a particle p moving on a circle of radius A with uniform angular speed ‘ω’ in anticlockwise direction. As the particle moves on the circumference of circle, its projection ‘M’ moves on the diameter xx¹ to and fro above centre ‘O’.

Suppose the particle starts from position ‘x’. After time‘t’ the position vector of the particle covers an angle of‘cot’ wive +ve X-axis as shown in figure. During this time the projection ‘M’ of the position vector of the particle on X-axis covers a displacement ‘x’.
From triangle OMP, We have
cosωt = \(\frac{\mathrm{x}}{\mathrm{~A}} \)
⇒  x = A cos ωt ———-(1)
The magnitude of acceleration of projection ‘M’ is equal to the component of centripetal acceleration of the particle, parallel to the diameter XX¹.
a_M = a_c cos ωt ———- (2)
a_M = Aω²cosωt [∵ centripetel acceleration a_c = Aω²]
From the figure acceleration of the projection M is opposite to the direction of its displacement ‘x’.
i.e., a_m = – Aω²cosωt ———(3)
a_m = -ω²x [∵ x = Acosωt]
∴ a_m α – x ———(4)
From this equation we conclude that in uniform circular motion the projec­tion of a particle on any diameter is in simple harpionic motion.

Question 3.
Show that the motion of a simple pendulum is simple harmonic and hence de­rive the equation for its time period. What is second’s pendulum?
Answer:
Consider a simple pendulum of length ‘l’ consists a bob of mass ‘m’ is suspended from rigid support‘s’.

The pendulum pulled through a side with small angular displacement ‘θ’ with released then it begins to oscillation about the mean position ‘θ’ at any instant of time, the pendulum at point A, the weight ‘mg’ resolved into two components, ‘mg cosθ’ balanced by the tension in a string, ‘mg sinG’ pulled the pendulum to its means position, hence ‘mg sin θ’ is called restoring force
F = – mg sinθ

If ‘θ’ is very small, then sinθ≈θ
F=-mg θ
From Newton’s IInd  law we have
F=ma
Substituting in the above equation

∴ The time period is independent of the amplitude of oscillation as long as the amplitudé is so small such that the bob oscillates along a straight line.

Second’s Pendulum :
The pendulum whose time period is two seconds is called seconds pendulum.

VII. Problems

Question 1.
A particle executes SHM such that, the maximum velocity during the oscillation is numerically equal to half of the maximum acceleration. What is the time period?
Answer:

Question 2.
A particle executing SHM has amplitude of 4 cm and its acceleration at a distance of 1 cm from the mean position is 3cms^-2. What will be the velocity, when it is at a distance of 2 cm from its mean position?
Answer:
Given
Amplitude, A=4 cm
Acceleration, a=3 cms^-2
Displacement, y=1 cm
Angular velocity

Question 3.
A 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 ?
Answer:
Given
Time period, T = 2s
time t =0.25 s

Question 4.
Calculate the change in the length of a simple pendulum of length 1 m, when its period of oscillation changes from 2s to 1.5s.
Answer:
For seconds pendulum T₁ = 2s
length l₂ = 1m
New time period T₂ = 1.5s
Length l₂ = ?

Question 5.
A simple pendulum in a stationary lift has time period T. What would be the effect on the time period when the lift
(i) moves up with uniform velocity
(ii) moves down with uniform velocity
(iii) moves up with uniform acceleration a
(iv) moves down with uniform acceleration ‘a’
(v) begins to fall freely under gravity?
Answer:
i) If the lift is moving up with uniform velocity then the effective acceleration is equal to g . There is no change in the time period.
ii) If the lift is moving down with uniform velocity then the effective acceleration is equal to g. There is no change in the time period.
iii) If the lift moves up with uniform acceleration ‘a’ then the effective acceleration is equal to g+a.
∴ T= \(2 \pi \sqrt{\frac{1}{\mathrm{~g}+\mathrm{a}}}\)
So time period decreases.
iv). If the lift moves down with uniform acceleration ‘ a ‘ then the effective acceleration is equal to g-a.
∴ T= \(2 \pi \sqrt{\frac{l}{\mathrm{~g}-\mathrm{a}}}\)
So time period increases.
v) If the lift begins to fall freely under gravity then effective acceleration is zero so time period is infinity.

Question 6.
A body describes simple harmonic motion with amplitude of 5 cm and a period of 0.2s. Find the acceleration and velocity of the body when the displacement is.
(a) 5 cm
(b) 3 cm
(c) 0 cm
Answer:
Given A=5 cm
T= 0.2s
ω=\(\frac{2 \pi}{\mathrm{~T}}= \frac{2 \pi}{0.2}\)=10π radian/sec

Question 7.
A freely falling body takes 2 seconds to reach the ground on a planet, when it is dropped from a height of 8 m. If the period of a simple pendulum is π seconds on the planet, calculate the length of the pendulum.
Answer:
Given
Height h =8m
Time taken to reach the ground,
t=2 s
But the time of fall of freely falling object is