A particle of mass m is executing SHM about its mean position. The total energy of the particle at given instant is

  • Option 1)

    \frac{\pi^{2}mA^{2}}{T^{2}}

  • Option 2)

    \frac{4\pi^{2}mA^{2}}{T^{2}}

  • Option 3)

    \frac{2\pi^{2}mA^{2}}{T^{2}}

  • Option 4)

    \frac{8\pi^{2}mA^{2}}{T^{2}}

 

Answers (1)

As we discussed in concept

Total energy in S.H.M. -

Total Energy = Kinetic + Potential Energy

- wherein

Total Energy =\frac{1}{2}K\left ( A^{2}-x^{2} \right )+\frac{1}{2}kx^{2}= \frac{1}{2}kA^{2}

Hence total energy in S.H.M. is constant

 

 Total energy = \frac{1}{2}mw^{2}A^{2}

T=\frac{1}{2}m\frac{4\pi ^{2}}{T^{2}}A^{2}

=\frac{2\pi ^{2}mA^{2}}{T^{2}}


Option 1)

\frac{\pi^{2}mA^{2}}{T^{2}}

Option is incorrect

Option 2)

\frac{4\pi^{2}mA^{2}}{T^{2}}

Option is incorrect

Option 3)

\frac{2\pi^{2}mA^{2}}{T^{2}}

Option is correct

Option 4)

\frac{8\pi^{2}mA^{2}}{T^{2}}

Option is incorrect

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