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A particle of a mass 10 gram is executing simple harmonic motion with an amplitude of 0.5m and periodic time of (\pi /5) seconds. The maximum value of the force acting on the particle is 

  • Option 1)

    25 \: N

  • Option 2)

    5 \: N

  • Option 3)

    2.5 \: N

  • Option 4)

    0.5 \: N

 

Answers (1)

best_answer

Maximum force

 \\*= m(a\omega^{2}) = ma\times\left[ \frac{4\pi^{2}}{T^{2}}\right ]\\* =0.5 \left [\frac{4\pi^{2}}{\frac{\pi^{2}}{25}} \right ] \times 0.01 \\*=0.5 N

 

Accelration in SHM -

A = \frac{dV}{dt} = a\omega ^{2}\cos \omega t

A = -\omega ^{2}y    [Formula]

- wherein

Rate of change of velocity

 

  


Option 1)

25 \: N

This is incorrect.

Option 2)

5 \: N

This is incorrect.

Option 3)

2.5 \: N

This is incorrect.

Option 4)

0.5 \: N

This is correct.

Posted by

Avinash

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