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  • Help me solve this The amplitude of a particle executing SHM with frequency of 60 Hz is 0.01 m. The maximum value of the accelration of the particle is

The amplitude of a particle executing SHM with frequency of 60 Hz is 0.01 m. The maximum value of the accelration of the particle is 

  • Option 1)

    144\pi ^{2}\; \frac{m}{s^{2}}

  • Option 2)

    144\: \frac{m}{s^{2}}

  • Option 3)

    \frac{144}{\pi^{2}}\; \frac{m}{s^{2}}

  • Option 4)

    288\pi ^{2}\; \frac{m}{s^{2}}

 

Answers (1)

Maximum acceleration

 \\*= a\omega^{2} = a\times4\pi^{2}n^{2} \\* =0.01 \times 4\times (\pi)^{2} \times (60)^{2} \\*=144\pi^{2}

 

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)

144\pi ^{2}\; \frac{m}{s^{2}}

This is correct.

Option 2)

144\: \frac{m}{s^{2}}

This is incorrect.

Option 3)

\frac{144}{\pi^{2}}\; \frac{m}{s^{2}}

This is incorrect.

Option 4)

288\pi ^{2}\; \frac{m}{s^{2}}

This is incorrect.

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subam

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