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  • A particle performs S.H.M. on x-axis with amplitude A and time period T. The time taken by the particle to travel adistance A/5 starting from rest is:Option: 1</

A particle performs S.H.M. on x-axis with amplitude A and time period T. The time taken by the particle to travel adistance A/5 starting from rest is:

Option: 1

\frac{T}{20}

 

 

 


Option: 2

\frac{T}{2\pi }\cos ^{-1}\left ( \frac{4}{5} \right )


Option: 3

\frac{T}{2\pi }\cos ^{-1}\left ( \frac{1}{5} \right )


Option: 4

\frac{T}{2\pi }\sin ^{-1}\left ( \frac{1}{5} \right )


Answers (1)

best_answer

 

Equation of S.H.M. -

a=-\frac{d^{2}x}{dt^{2}}= -w^{2}x

w= \sqrt{\frac{k}{m}}

 

- wherein

x= A\sin \left ( wt+\delta \right )

 

 

Method 1 :

Particle is starting from rest ,ie from one of its extreme position

As particle moves a distance \frac{A}{5},We can represent it on a circle as shown

\cos \theta = \frac{4A/5}{A}= \frac{4}{5}\theta = \cos ^{-1}\left ( \frac{4}{5} \right )

\omega t= \cos ^{-1}\left ( \frac{4}{5} \right )

t=\frac{1}{\omega }\cos ^{-1}\left ( \frac{4}{5} \right )=\frac{T}{2\pi }\cos ^{-1}\left ( \frac{4}{5} \right )

Method 2:

As starts from rest i.e. from extreme position

X= A\sin \left ( \omega t+\phi \right )

At \ t= 0; X=A\Rightarrow \phi = \frac{\pi }{2}

\therefore A-\frac{A}{5}= A\cos \omega t\Rightarrow \omega t= \cos ^{-1}\frac{4}{5}

\Rightarrow t= \frac{T}{2\pi }\cos ^{-1}\left ( \frac{4}{5} \right )

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