Question

A particle is executing S.H.M. with amplitude A and Time period T. Time taken by the particle to reach from extreme position to A/2


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Answer 1

Here is the answer....

Good luck..

Answer 2

The time taken by the particle is $\dfrac{T}{12}$

Explanation:

Given that,

Displacement $y_{max}= \dfrac{A}{2}$

Amplitude = A

Time period = T

A particle is executing S.H.M. with amplitude A and Time period T.

We need to calculate the time taken

Using the general equation of simple harmonic motion

$y_{max}=a\sin\omega t$

Where, a = amplitude

t = time taken

$\omega$ = angular frequency

Put the value in the equation

$\dfrac{A}{2}=A\sin\times\dfrac{2\pi}{T}\times t$

$\dfrac{2\pi}{T}\times t=\sin^{-1}(\dfrac{1}{2})$

$t=30\times\dfrac{T}{2\pi}$

$t=\dfrac{15 T}{\pi}$

$t=\dfrac{T}{12}$

Hence, The time taken by the particle is $\dfrac{T}{12}$

Learn more :

Topic : simple harmonic motion

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