A particle of mass m executes simple harmonic motion with amplitude a  and frequency  \upsilon. The average kinetic energy during its motion from the  position of equilibrium to the end is

  • Option 1)

    2\pi ^{2}ma^{2}\upsilon ^{2}

  • Option 2)

    \pi ^{2}ma^{2}\upsilon ^{2}

  • Option 3)

    \frac{1}{4}ma^{2}\upsilon ^{2}

  • Option 4)

    4\pi ^{2}ma^{2}\upsilon ^{2}

 

Answers (1)
V Vakul

As we learnt in

Potential energy in S.H.M. -

P.E.= \frac{1}{2}Kx^{2}

 

- wherein

Where K= m\omega ^{2}

 

 

 

For a particle to execute simple harmonic motion its displacement at any time  t is given by  x(t)=a(cos\, \omega t+\phi )

where, a= amplitude,\omega =angular \; frequency,\phi =phase \; constant.

Let \; us \; choose\: \phi =0

\therefore \; \; x(t)=a\: cos\: \omega t

Velocity of a particle \nu = \frac{dx}{dt}= -a\: \omega \: \sin \omega t

Kinetic \: energy\: of \: a \: particle\: is\: K= \frac{1}{2} m\nu ^{2}

                                                                        = \frac{1}{2}ma^{2}\omega ^{2}\sin ^{2}\omega t

Average\: kinetic \: energy\: < K>

= < \frac{1}{2}ma^{2}\omega ^{2}\sin ^{2}\omega t>

= \frac{1}{2}m\omega ^{2}a^{2}< \sin ^{2}\omega t>

= \frac{1}{2}m\omega ^{2}a^{2}\left ( \frac{1}{2} \right )

\left [ \because \sin ^{2}\Theta > = \frac{1}{2} \right ]

= \frac{1}{4}ma^{2}\left ( 2\pi \upsilon \right )^{2}\: \: \: \: \: \: \: \left [\because \omega = 2\pi \upsilon \right ]

= \pi ^{2}ma^{2}\upsilon

Correct option is 2.

 


Option 1)

2\pi ^{2}ma^{2}\upsilon ^{2}

This is an incorrect option.

Option 2)

\pi ^{2}ma^{2}\upsilon ^{2}

This is the correct option.

Option 3)

\frac{1}{4}ma^{2}\upsilon ^{2}

This is an incorrect option.

Option 4)

4\pi ^{2}ma^{2}\upsilon ^{2}

This is an incorrect option.

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