The maximum velocity of a particle, executing simple harmonic motion with an amplitude 7 mm, is 4.4 m/s. The period of oscillation is

  • Option 1)

    100 s

  • Option 2)

    0.01 s

  • Option 3)

    10 s

  • Option 4)

    0,1 s

 

Answers (1)

As we learnt in

Relation of velocity and displacement -

v= w\sqrt{A^{2}-x^{2}}
 

- wherein

\rightarrow  x is displacement from mean position

\rightarrow  A is Amplitude.

 

 

 

:    Maximum velocity  \nu _{m}= a\omega = a\left ( \frac{2\pi }{T} \right )

\therefore \: \: \: T= \frac{2\pi a}{\nu _{m}}= 2\times \frac{22}{7}\times \frac{\left ( 7\times 10^{-3} \right )}{4.4}

=10^{-2} sec= 0.01sec.

Correct option is 2.


Option 1)

100 s

This is an incorrect option.

Option 2)

0.01 s

This is the correct option.

Option 3)

10 s

This is an incorrect option.

Option 4)

0,1 s

This is an incorrect option.

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