Simple Harmonic Motion Test

Result

Simple Harmonic Motion Test - 10

Score
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out of -
Rank
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TIME Taken - -

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Question 1
1 / -0

A horizontal plank has a rectangular block placed on it. The plank starts oscillating vertically and simple harmonically with an amplitude of 40 cm. The block just loses contact with the plank when the later is momentarily at rest. Then

A
the period of oscillation is 2π/3

B
the block weighs double its weight when the plank is at one of the positions of momentary at rest.

C
the block weighs 1.5 times its weights on the plank halfway down

D
the block weighs more than its true weight on the plank when the latter moves fastest

Solution

At one of the extreme positions, weight of block=restoring force. At the other extreme position, the weight of block and restoring force both act downward direction. So the force on the block is doubled than its weight.
Question 2
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A body has a time period T₁ under the action of one force and T₂ under the action of another force, the square of the time period when both the forces are acting simultaneously in the same direction is

A

B

C

D

Solution
Question 3
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Motion of an oscillating liquid column in a U-tube is:

A
periodic but not simple harmonic

B
non-periodic

C
simple harmonic and time period is independent of the density of the liquid

D
simple harmonic and time-period is directly proportional to the density of the liquid

Solution

The motion of an oscillating liquid column in a U-tube is SHM with the period, , where l is the height of the liquid column in one arm of U-tube in equilibrium position of liquid. Therefore, T is independent of the density of the liquid.
Question 4
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This time period of a particle undergoing SHM is 16s. It starts motion from the mean position. After 2s, velocity is 0.4 ms⁻¹. The amplitude is:

A
1.44m

B
0.72m

C
2.88m

D
0.36m

Solution
Question 5
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A particle is performing simple harmonic motion along the x-axis with amplitude 4 cm and time period 1.2 s. The minimum time taken by the particle to move from x = + 2 cm to x = + 4 cm and back again is given by:

A
0.4 sec

B
0.3 sec

C
0.2 sec

D
0.6 sec

Solution

When the particle is at x=2, the displacement is y=4-2=2cm. If t is the time taken by the particle to go from x=4cm to x=2cm, then y=

The time taken to move from x=+2 cm to x=+4 cm and back again=2t=2x0.2 sec=0.4 sec.
Question 6
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The acceleration of a particle performing SHM is 12 cm sec⁻² at a distance of 3 cm from the mean position. Its time period is:

A
2.0s

B
3.14s

C
0.5s

D
1.0s

Solution

or
Question 7
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The acceleration d²x/dt² of a particle varies with displacement x as d²x/dt²= −kx

where k is a constant of the motion. The time period T of the motion is equal to :

A
2πk

B
2π√k

C
2π / √k

D
2π/k

Solution
Question 8
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A coin is placed on a horizontal platform, which undergoes horizontal SHM about a mean position O. The coin placed on the platform does not slip, the coefficient of friction between the coin and the platform is μ. The amplitude of oscillation is gradually increased. The coil will begin to slip on the platform for the first time:

A
at the mean position

B
at the extreme position of oscillations

C
for an amplitude of μg/ω²

D
for an amplitude of g/μω²

Solution

Let O be the position and x be the distance of coin from O. The coil will slip if the pseudo force on a coin just becomes equal to force of friction i.e., ma = mxω²= μmg

The coin will slip if, x=maximum amplitude A

mAω²= μ mg

or A = μg/ω²
Question 9
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A particle moves according to the law, x = r cos πt/2 . The distance covered by it in the time interval between t=0 to t=3 s is:

A
r

B
2r

C
3r

D
4r

Solution

∴ In 3s, the particle goes from one extreme to another extreme and then back to the mean position. So, the distance travelled=2r+r=3r
Question 10
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A particle is executing SHM of period 24 sec and of amplitude 41 cm with O as equilibrium position. The minimum time in seconds taken by the particle to go from P to Q, where OP=-9cm and OQ=40cm is:

A
5

B
6

C
7

D
9

Solution

For displacement OQ=40cm; let t₁ be the time taken, then-

On solving, t₁= 5.16 s

For displacement OQ=-9cm, let t₂ be the time taken, then-

On solving t₂=0.84s

Total time=5.16+0.84=6.00s