Question

The displacement of a particle moving in S.H.M. at any instant is given by y = a sin(w t). The
acceleration after time t= T/4 is (where T is the time period)​

Answer 1

Explanation:

pehle y=a sin (wt)ko differentiate karo...

to velocity milegi..

fur us velocity ko differentiate karo..

to acc. milega..

usme t=T/4 rakh do..

ans mol jayega..

...I hope it helps....

Answer 2

The acceleration is $-(\omega^2)a$

Explanation:

Given that,

Displacement equation is

$y=a\sin(\omega t)$

Time $t =\dfrac{T}{4}$

We need to calculate the acceleration for S.H.M

Using equation of S.H.M

$y=a\sin(\omega t)$

On differentiating

$\dfrac{dy}{dt}=a\omega\cos(\omega t)$

Again on differentiating

$\dfrac{d^2y}{dt^2}=-a\omega^2\sin(\omega t)$

$a(t)=-\omega^2y(t)$

Put the value into the formula

$a(\dfrac{T}{4})=-(\omega^2)a\sin(\dfrac{2\pi}{T}\times\dfrac{T}{4})$

$a(\dfrac{T}{4})=-(\omega^2)a$

Hence, The acceleration is $-(\omega^2)a$

Learn more :

Topic : Simple harmonic motion

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