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A doubt on Waves

Here is a question,

A string of mass \(m\) and length \(l\), both fixed at both ends is vibrating in its fundamental mode. The maximum amplitude is \(a\) and the tension in the string is \(T\). Find the energy of vibrations of the string in terms of \(a\), \(T\) and \(l\).

This is how I answered,

Let the equation of wave be \(y=f\left( x,t \right) =a\sin { \left( \omega t-kx \right) } \)

Differentiating w.r.t \(t\),

\(\frac { \delta y }{ \delta t } ={ v }_{ p }=a\omega \cos { \left( \omega t-kx \right) } \)

\(\therefore \) Kinetic Energy, \(K.E=\frac { 1 }{ 2 } m{ a }^{ 2 }{ \omega }^{ 2 }\cos ^{ 2 }{ \left( \omega t-kx \right) } \)

\(\therefore\)Average Kinetic Energy, \(\left< K.E \right> =\frac { \int _{ 0 }^{ T }{ \frac { 1 }{ 2 } m{ a }^{ 2 }{ \omega }^{ 2 }\cos ^{ 2 }{ \left( \omega t-kx \right) } dt } }{ \int _{ 0 }^{ T }{ dt } } =\frac { 1 }{ 4 } m{ a }^{ 2 }{ \omega }^{ 2 }\)

Potential Energy, \(P.E=\frac { 1 }{ 2 } m{ \omega }^{ 2 }{ a }^{ 2 }\sin ^{ 2 }{ \left( \omega t-kx \right) } \)

\(\therefore\) Average Potential Energy,\(\left< P.E \right> =\frac { \int _{ 0 }^{ T }{ \frac { 1 }{ 2 } m{ a }^{ 2 }{ \omega }^{ 2 }\sin ^{ 2 }{ \left( \omega t-kx \right) } dt } }{ \int _{ 0 }^{ T }{ dt } } =\frac { 1 }{ 4 } m{ a }^{ 2 }{ \omega }^{ 2 }\)

Therefore Total energy=\(\left< K.E \right>+\left< P.E \right>=\frac { 1 }{ 2 } m{ a }^{ 2 }{ \omega }^{ 2 }\)

Now,

\({ \omega }^{ 2 }={ v }^{ 2 }{ k }^{ 2 }={ v }^{ 2 }\frac { 4{ \pi }^{ 2 } }{ { \lambda }^{ 2 } } =\frac { T }{ \frac { m }{ L } } \frac { 4{ \pi }^{ 2 } }{ { \lambda }^{ 2 } } =\frac { 2T{ \pi }^{ 2 } }{ m\lambda } =\frac { T{ \pi }^{ 2 } }{ mL } \)

\(\therefore\) Total Energy\(=\boxed{\frac { T{ a }^{ 2 }{ \pi }^{ 2 } }{ 2L } }\)

But the given answer is \(\boxed{\frac { T{ a }^{ 2 }{ \pi }^{ 2 } }{ 4L} }\)

Can someone explain this?

Note by Anandhu Raj
1 year, 8 months ago

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1 vote

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I think you're right.
See this.
Unless, they meant only the energy due to the motion, which is the kinetic energy.

Ameya Daigavane - 1 year, 8 months ago

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Again you are there to help me! Thank you, Sir :)

Anandhu Raj - 1 year, 8 months ago

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