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Question

A particle of mass 10g is executing simple harmonic motion with an amplitude of 0.5m and periodic time of π5s. The maximum value of the force acting on the particle is

A
25 N
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B
5 N
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C
2.5N
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D
0.5 N
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Solution

The correct option is D 0.5 N.

Given,

Mass = m = 10 g

Amplitude = A = 0.5 m

Time period = T = π5 s

From, F=ma we know that for constant m the force experienced by the body will be maximum when acceleration a is maximum.

For a particle executing simple harmonic motion the acceleration at a displacement x is given by,

a=ω2x

where ω=2πT is the angular frequency of the SHM.

Hence acceleration will be maximum at the maximum displacement i.e.; at x = A.

Fmax=mω2A=m4π2T2A
Fmax=101000×4×(π)2(π5)2×0.5=0.5N.


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