250+ TOP MCQs on Oscillations – Systems Executing Simple Harmonic Motion | Class 11 Physics

3. The springs in the given diagram are fixed to the walls. If the block is displaced by a distance x from the mean position, find the time period of oscillations in seconds. Assume that there is no friction.

a) 2π √(2m/3k)
b) 2π √(7m/3k)
c) 2π √(2m/7k)
d) 2π √(2m/5k)
Answer: c
Clarification: When the block is displaced by x in one direction, let’s say left side, all the springs will exert a force on the block towards right. The two springs will be compressed by length x and the right one will be extended by length x.
Net force on block = kx + 2kx + (k/2)x = 3.5kx.
Displacement is towards left and force is towards right, so we write F = -3.5kx.
F = ma = -3.5kx.
∴ a = -(3.5k/m)x.
We know that in SHM a = -w²x.
∴ w = √(3.5k/m).
∴ T = 2π/w = 2π √(m/3.5k) = 2π √(2m/7k) s.

4. If the block is displaced by ‘x’ from the mean position, what will be the time period of oscillations?

a) 2π √(2m/3k) s
b) 2π √(3m/k) s
c) 2π √(m/k) s
d) 2π √(4m/k) s
Answer: b
Clarification: Let spring with constant k be spring 1 and that with constant k/2 be spring 2.
For series connection, effective spring constant is given by (1/k + 2/k)^-1 = k/3.
Thus, time period = 2π √ (3m/k) s.

5. Find the effective spring constant.

a) 8/5
b) 8/31
c) 5/8
d) 31/8
Answer: a
Clarification: The springs with spring constant 5 and 3 are parallel,
so their effective spring constant is 5+3 = 8.
Now, spring with spring constant 2 is in series with effective constant 8,
so net spring constant
= (1/8 + 1/2)^-1 = 8/5.

6. A spring mass system is on a horizontal plane. Spring constant is 2N/m & mass of block is 2kg. The spring is compressed by 10cm and released. A small ball is released from a height of 2m above the block on the horizontal plane. At what time should the ball be left if the block is released from the extreme position at t=0, given that the ball has to fall on the block when the speed of block is maximum?
a) 1.57s
b) 0.94s
c) 0s
d) 0.63s
Answer: b
Clarification: w = √2/2 = 1. Time period of SHM = 2π/w = 2π s.
Speed of the block will be maximum at the mean position.
Time to reach mean position from extreme position = T/4
= 2π/4 = π/2 =1.57s.
Time for the small ball to fall down = √(2h/g) = √(4/9.8) = 0.63s.
∴ Time when it should be released
= 1.57 – 0.63 s
= 0.94s.

7. A spring mass system is hanging from the ceiling of a lift. It has been like this for quite some time. Suddenly the cables holding the lift are cut. What will be the amplitude of oscillation of the block when observed by a person standing (of mass M) in the lift?
a) mg/k
b) (M-m)g/k
c) The observer will not observe a SHM
d) 2mg/k
Answer: a
Clarification: When the system was stationary, the extension in the spring will be x_o such that kx_o= mg or x₀ = mg/k.
When the cables are cut, the entire system will be in free fall.
The observer will add a pseudo force on the block upwards and equal to mg.
Now the spring has an extension of mg/k
and the restoring force by the spring will be corresponding to this extension,
so the amplitude of SHM for the observer is mg/k.

8. A spring block system is on a horizontal surface. The block is compressed by 10cm. What should be the coefficient of friction between the block and ground such that the block comes to a stop at the mean position? Mass of block = 0.5kg & spring constant = 3N/m.
a) 0.03
b) 0.06
c) 0.12
d) 0.05
Answer: b
Clarification: When the block is released, the spring force will act towards the mean position and friction force will act against velocity. The work done by these forces should be equal to potential energy of the system in the compressed state. ∴ Work done in dx displacement = (μmg – kx)dx.
On integrating this, with dx varying from 0 to 0.1, we get
Work = (μmgx – kx√/2)₀^0.1= 0.5μ – 0.015.
This should be equal to initial potential energy of the system.
0.5μ – 0.015 = 1/2kA√
∴ 0.5μ – 0.015 = 1/2 * 3 * 0.01
∴ 0.5μ – 0.015 = 0.015
∴ 0.5μ = 0.03 OR μ = 0.06.

9. Two springs in horizontal spring mass systems have spring constants in the ratio 2:3. By what ratio should they be extended from their mean positions so that they have the same value of maximum speed? The masses of blocks are the same in both cases.
a) 3/2
b) √3 / √2
c) 9/4
d) 4/9
Answer: b
Clarification: w = √(k/m), Maximum speed = Aw.
According to question: A₁w₁ = A₂w₂.
Their extensions from mean positions will be their amplitudes.
A₁/A₂ = w₂/w₁
= √k₂/√k₁
= √3 / √2.

10. A spring is fixed to the ground as shown below. A block of mass m is fixed to the spring. The spring is compressed by 5cm. Imagine that we can remove the spring suddenly and the block will fly off with the speed it has at that instant. The spring has to be removed at such a position that the distance travelled upwards by the block relative to that position is maximum. Find that position.

a) Mean position
b) 2cm above mean position
c) 4cm above mean position
d) 2cm below mean position
Answer: a
Clarification: The given condition can be achieved when the spring is made to disappear at mean position because the block will have maximum speed.

11. Which of the following factors affects the time period of a simple pendulum?
a) Mass of bob
b) Shape of bob
c) Length of string
d) Angle of release from 0° to 20°
Answer: c
Clarification: We neglect air resistance, so the shape of the bob will not matter.
Time period is given by:
2π √(l/g).
Thus, it depends on the length of the pendulum and value of g.

12. A simple pendulum,hanging from the ceiling, is released from an angle . What is the restoring force?

a) mg
b) mgcosθ
c) mgsinθ
d) mgtanθ
Answer: c
Clarification: The component of mg which will cause change in speed is mgsinθ.
This force is always towards the mean position and is a function of θ.

13. What is the length of a seconds pendulum?
a) 1m
b) 2m
c) 9.8m
d) 0.5m
Answer: a
Clarification: Time period of seconds pendulum = 2s.
∴ 2 = 2π √(L/9.8)
OR L = 1m.

14. Two simple pendulums of the same length are released from 10° & 20° respectively. Select the correct statement.
a) The second pendulum have will have a double time period than that of the first
b) They both will reach the mean position at the same time
c) The first pendulum will have a double time period than that of the second
d) Their maximum speed will be the same
Answer: b
Clarification: Their time period will be the same as length is the same. So time to reach the mean position from their respective extremes will be the same. Therefore they will reach the mean position at the same time. They both have different potential energies at the start, so their maximum speeds will be different. Another way to look at this is that the speed of the 2nd pendulum should be more as it has to cover greater arc length in the same amount of time.

15. What will happen to the time period of a simple pendulum if temperature of the surroundings is significantly increased?
a) Increase
b) Decrease
c) Remain the same
d) The oscillations come to a stop
Answer: a
Clarification: The time period of a simple pendulum is given by: T = 2π √(l/g).
If the temperature increases,
length of string will increase and thus,
the time period will increase.

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