Question

A simple pendulum performs S.H.M of
period 4 seconds. How much time after
crossing the mean position, will the
displacement of the bob be one third of
its amplitude.
Ans: 0.2163 s]

Answer 1

Given that,

A simple pendulum performs S.H.M of  period 4 seconds.

To find,

How much time after  crossing the mean position, will the  displacement of the bob be one third of  its amplitude.

Solution,

Let after time t it crosses (1/3)rd of its amplitude. So, the equation of SHM is given by :

$x=A\sin\omega t$

Here, x = A/3

So,

$\dfrac{A}{3}=A\sin\omega t\\\\\sin\omega t = \dfrac{1}{3}\\\\\omega t=\sin^{-1}(\dfrac{1}{3})$

Since, $\omega=\dfrac{2\pi }{T}$

So,

$\omega t=\sin^{-1}(\dfrac{1}{3})\\\\(\dfrac{2\pi}{T})t=0.3398\\\\t=\dfrac{0.3398T}{2\pi}\\\\t=\dfrac{4\times 0.3398}{2\pi}\\\\t=0.2163\ s$

So, the time is 0.2163 seconds.

Answer 2

Answer:

0.2163s

Explanation:

Given:

Time period (T)=4sec

To find:

The time required for the particle to displace (1÷3)rd  of the amplitude..

The displacement equired of the particle is given by x=Asinωt Initially the particle is at the mean position so t=0 and x=0.

Let after time t is crosses (1÷3)rd of its amplitude x=Asinωt

(A÷3)=Asinωt

sinωt=1÷3

By applying the value of sin^-(1÷3)ωt=0.3398

(2π)t=0.3398

t=0.3398×4÷2π

t=0.216sec