Question

The wave y1 = 0.12 sin π/3(5x+200t-1) propagates on a string of linear density 0.02 kg/m.

Answer 1

The average power transmitted by string wave is given by the expression $P=\frac{1}{2} \mu\omega^2A^2v$, Where μ  is a linear density of a string, ω is angular frequency of the wave, A is an amplitude of the wave, v  is a speed of wave.

Since $y(x,t)= A sin (kx+\omega t-\psi )$  comparing with  $y_1 = 0.12 sin \frac{\pi}{3}(5x+200t-1)$

 A = 0.12 m, ω = $\frac{200\pi }{3} s^{-1}$, k = $\frac{5\pi }{3} m^{-1}$

Speed of wave, $v = \frac{\omega }{k} = \frac{\frac{200\pi }{3}}{\frac{5\pi }{3} } =40m/s$

So power $P = \frac{1}{2} *0.02*(\frac{200\pi }{3} )^2*0.12^2*40=253 W$