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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 $$l$$.

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, 10 months ago

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Are vah yyaar maja hi aa Gaya shandaar

- 1 week ago

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

- 1 year, 10 months ago

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

- 1 year, 10 months ago