Question

It is found that a 2.9 m segment of a long
string contains 3 complete wavelengths and
has a mass of 190 g. At one point it is vibrating
sinusoidally with a frequency of 46 Hz and
a peak to valley displacement of 0.66 m.
What power is being transmitted along the
string?
Answer in units of W.

Answer 1

form the given data, 3*lamda = 2.9

==> lamda = 2.9/3

= 0.967 m

Average power, P = 0.5*v*mue*w^2*A^2

v(wave speed) = lamda*f

= 0.967*46

= 44.47 m/s

mue(mass per unit length) = M/L

= 0.19/2.9

= 0.0655 kg/m

w(angular frequency) = 2*pi*f

= 2*pi*46

= 288.88 rad/s

A(amplitude) = 0.66 m

so, P = 0.5*44.47*0.0655*288.88^2*0.66^2

= 52942 Watts