Question

The transverse displacement of a string (mass per length m)
in which a wave is travelling with speed v is,
y = Ym sin(kx - wt
The average rate at which kinetic energy is transported
in the string is​

Answer 1

Answer:

$\frac{d(KE)}{dt} =\frac{1}{4}\mu A^2\omega^2 v$

Explanation:

Consider an element in a wave of length dx and mass dm,

Let the string have uniform mass density μ.

∴$\mu=\frac{dm}{dx}$

Since the particle which we considered is fixed along the direction of propagation, but can oscillate perpendicular to the direction of propagation of the wave.

Here the particle oscillates harmonically, i.e, it exhibits S.H.M

So KE of the particle in S.H.M is given by:-

$=>d(KE) =\frac{1}{2}(dm)\omega^2 A^2cos^2(kx-\omega t) \\\\=>KE=\int\limits^\lambda_0 {\frac{1}{2}(\mu)(dx)\omega^2 A^2cos^2(kx-\omega t)}$

On integrating, you get KE for one wavelength,

$\boxed{KE =\frac{1}{4}\mu A^2\omega^2 \lambda}$

Now if we differentiate this expression with time, we get rate of change/ transfer of KE:

$\frac{d(KE)}{dt} =\frac{d}{dt}(\frac{1}{4}\mu A^2\omega^2 \lambda)$

Since $\boxed{\frac{d\lambda}{dt}=v_w_a_v_e }$

$=>\boxed{\boxed{\frac{d(KE)}{dt} =\frac{1}{4}\mu A^2\omega^2 v}}$

Hope this answer helped you :)