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  • Answer please! A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration.then its periodic time in seconds is :

A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration.then its periodic time in seconds is :

 

  • Option 1)

    \frac{3}{8}\pi

  • Option 2)

    \frac{4\pi}{3}

  • Option 3)

    \frac{8\pi}{3}

  • Option 4)

    \frac{7}{3}\pi

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 )

 

 

Relation of velocity and displacement -

v= w\sqrt{A^{2}-x^{2}}
 

- wherein

\rightarrow  x is displacement from mean position

\rightarrow  A is Amplitude.

Given

\omega \sqrt{A^{2}-X^{2}} =\omega ^{2}X

\omega \times 4 =3

\frac{2\pi }{T}\times 4 =3

T = \frac{8\pi }{3}sec

 

 

 

 


Option 1)

\frac{3}{8}\pi

Option 2)

\frac{4\pi}{3}

Option 3)

\frac{8\pi}{3}

Option 4)

\frac{7}{3}\pi

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