Simple Harmonic Motion Test

Result

Simple Harmonic Motion Test - 21

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Question 1
1 / -0

Two identical springs have the same force constant 73.5 Nm^-1. The elongation produced in each spring in three cases shown in Figure-1, Figure- 2 and Figure- 3 are: (g = 9.8 ms^-2)

A

B

C

D

Solution
Question 2
1 / -0

A particle of mass 10 g is undergoing SHM of amplitude 10 cm and period 0.1 sec. The maximum value of the force on the particle is about:

A
5.6 N

B
2.75 N

C
3.5 N

D
4 N

Solution
Question 3
1 / -0

Which of the following graphs best represents the variation of acceleration 'a' of a particle executing SHM with displacement x?

A

B

C

D

Solution

The acceleration of a particle performing SHM is given by:

a = −ω²x

ω²

is always positive.

So the equation represents a straight line passing through origin with a negative slope.
Question 4
1 / -0

Two identical pendulums oscillate with a constant phase difference π/4 and the same amplitude. If the maximum velocity of one is v, the maximum velocity of the other will be:

A
v

B
√2v

C
2v

D

Solution

The maximum velocity achieved by a pendulum is independent of the phase difference between the two pendulums.
Question 5
1 / -0

A simple pendulum suspended from the ceiling of a stationary lift has period T₀. When the lift descends at a steady speed, the period is T₁, and when it descends with constant downward acceleration, the period is T₂. Which one of the following is true?

A
T₀= T₁= T₂

B
T₀= T₁< T₂

C
T₀= T₁> T₂

D
T₀< T₁< T₂

Solution

The time period is given by:

g_eff= g when the lift is stationary or moving with constant speed.

So, T_o = T₁

g_eff = g−a when it descends with a constant downward acceleration a.
Question 6
1 / -0

If a Second's pendulum is moved to a planet where the acceleration due to gravity is 4 times the acceleration due to gravity on earth, to keep the time period of the second's pendulum same, the length of the second's pendulum on the planet should be made:

A
2 times

B
4 times

C
8 times

D
15 times

Solution

New acceleration due to gravity is given by, g' = 4g

Since the time period of a simple pendulum is given by

So,
Question 7
1 / -0

Two pendulums of lengths 1.21 m and 1.0 m start vibrating. At some instant, the two are at the mean position in the same phase. After how many vibrations of the longer pendulum, the two will be in the phase again?

A
10

B
11

C
20

D
21

Solution

Since the time period depends on the square root of the length,
Question 8
1 / -0

The time period of oscillations of a simple pendulum is 1 minute. If its length is increased by 44%, then its new time period of oscillations will be:

A
96 s

B
58 s

C
82 s

D
72 s

Solution
Question 9
1 / -0

A simple pendulum is oscillating in a trolley moving on a horizontal straight road with constant acceleration a. If the direction of motion of the trolley is taken as positive x-direction and the vertically upward direction as positive y-direction, then the mean position of the pendulum makes an angle:

A
with y-axis in +x direction

B
with y-axis in -x direction

C
with y-axis in +x direction

D
with y-axis in -x direction

Solution

At the mean position:
Question 10
1 / -0

The time period of oscillations of a second's pendulum on the surface of a planet having mass and radius doubled of those of earth is:

A
4 s

B
1 s

C
√2 s

D
2√2 s

Solution

The time period is given by,