Question

a string of linear mass density 0.8 kg/m is stretched to a tension 500N. the mean power required to maintain a traveling wave of amplitude of 10 mm and wavelength 0.5 m is
(a) 70W (b) 85.3 W
(c) 98.7W (d) 110 W

Answer 1

we know, power transmitted in a string is given by, $P=\frac{1}{2}\mu\omega^2A^2\nu$

here, $\mu$ is linear mass density.

A is amplitude, $\nu$ velocity of wave propagation in string.

$\omega$ is angular frequency.

we should use formula, $\omega=\frac{2π\nu}{\lambda}$

here, $\lambda$ is wavelength.

so, now, power formula is $P=\frac{1}{2}\frac{4\pi^2A^2\nu^3}{\lambda^2}$

here , A = 10mm = 0.01m , $\lambda$ = 0.5m

using formula to find velocity of sound,

$\nu=\sqrt{\frac{T}{\mu}}$

here, T = 500N , $\mu$ = 0.8 kg/m

so, $\nu=\sqrt{\frac{500}{0.8}}=25$

now, power , P = 1/2 × {(4π² × (0.01)² × (25)³}/(0.5)²

= 98.69 ≈ 98.7 W

hence, option (C) is correct.