A doubt on Waves

Here is a question,

A string of mass mm and length ll, both fixed at both ends is vibrating in its fundamental mode. The maximum amplitude is aa and the tension in the string is TT. Find the energy of vibrations of the string in terms of aa, TT and ll.

This is how I answered,

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

Differentiating w.r.t tt,

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

\therefore Kinetic Energy, K.E=12ma2ω2cos2(ωtkx)K.E=\frac { 1 }{ 2 } m{ a }^{ 2 }{ \omega }^{ 2 }\cos ^{ 2 }{ \left( \omega t-kx \right) }

\thereforeAverage Kinetic Energy, <K.E>=0T12ma2ω2cos2(ωtkx)dt0Tdt=14ma2ω2\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=12mω2a2sin2(ωtkx)P.E=\frac { 1 }{ 2 } m{ \omega }^{ 2 }{ a }^{ 2 }\sin ^{ 2 }{ \left( \omega t-kx \right) }

\therefore Average Potential Energy,<P.E>=0T12ma2ω2sin2(ωtkx)dt0Tdt=14ma2ω2\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=<K.E>+<P.E>=12ma2ω2\left< K.E \right>+\left< P.E \right>=\frac { 1 }{ 2 } m{ a }^{ 2 }{ \omega }^{ 2 }

Now,

ω2=v2k2=v24π2λ2=TmL4π2λ2=2Tπ2mλ=Tπ2mL{ \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=Ta2π22L=\boxed{\frac { T{ a }^{ 2 }{ \pi }^{ 2 } }{ 2L } }

But the given answer is Ta2π24L\boxed{\frac { T{ a }^{ 2 }{ \pi }^{ 2 } }{ 4L} }

Can someone explain this?

Note by Anandhu Raj
3 years, 8 months ago

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1 vote

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I think it will be a standing wave and not a progressive wave, whose equation you have written. Hence, amplitude for every point of the string is different.

Arka Das - 10 months ago

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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 - 3 years, 8 months ago

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

Anandhu Raj - 3 years, 8 months ago

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Are vah yyaar maja hi aa Gaya shandaar

Sagar Chaturvedi - 1 year, 10 months ago

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