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  • A particle is performing simple harmonic motion along the  with an amplitude <img alt="6 \mathrm{~cm}" src

A particle is performing simple harmonic motion along the x-axis with an amplitude 6 \mathrm{~cm} and time period 1.2\, \, \mathrm{sec}. The minimum time taken by the particle to move from x=3 \mathrm{~cm} to \mathrm{x}=+6 \mathrm{~cm} and back again is given by

Option: 1

2.4 \mathrm{~s}


Option: 2

0.4 \mathrm{~s}


Option: 3

4.6 \mathrm{~s}


Option: 4

5.9 \mathrm{~s}


Answers (1)

best_answer

Time taken by particle to move from  \mathrm{x}=0  (mean position) to \mathrm{x}=6 (extreme position), 

=\frac{T}{4}=\frac{1.2}{4}=0.3 \mathrm{sec}

Let t be the time taken by the particle to move from  \mathrm{x}=0 \text { to } \mathrm{x}=3 \mathrm{~cm} \text {. }

Using the equation for simple harmonic motion, we have:

3=6 \sin \left(\frac{2 \pi}{T} t\right)

Simplifying the equation, we get:

\sin \left(\frac{2 \pi}{1.2} t\right)=\frac{1}{2}

Taking the inverse sine on both sides, we have:

\frac{2 \pi}{1.2} t=\sin ^{-1} 1\left(\frac{1}{2}\right)

Solving for \mathrm{t} , we get:

t=\frac{1.2}{2 \pi} \sin ^{-1}\left(\frac{1}{2}\right)=\frac{1.2}{2 \pi} \times \frac{\pi}{6}=0.1 \mathrm{sec}

Now, the time taken to move from \mathrm{x}=3 \mathrm{~cm} to \mathrm{x}=6 \mathrm{~cm} will be 0.3 \mathrm{sec}-0.1 \mathrm{sec}=0.2 \mathrm{sec} Therefore, the total time taken to move from \mathrm{x}=3 \mathrm{~cm} to \mathrm{x}=6 \mathrm{~cm} and back again is 2 \mathrm{x}$ $0.2 \mathrm{sec}=0.4 \mathrm{sec}

Posted by

avinash.dongre

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