Question

A 1.88-m long rope has a mass of 0.139 kg. The tension is 55.4 N. An oscillator at one end sends a harmonic wave with an amplitude of 1.09 cm down the rope. The other end of the rope is terminated so all of the energy of the wave is absorbed and none is reflected. What is the frequency of the oscillator if the power transmitted is 118 W?

Answer 1

You need to know that the power in a vibrating string is
P = (1/2) ì ù^2 A^2 v .

Reference: http://hyperphysics.phy-astr.gsu.edu/hbase/waves/powstr.html

In that equation, ì (mu) is the mass per unit length, ù (omega) is the angular frequency of oscillation, A is the amplitude and v is the wave speed,

v = sqrt (T/ì). T is the tension.

With those equations, you can solve the problem yourself. Your value of mu (ì) is
ì = 0.0739 kg/m.
The wave speed is
v = sqrt(55.4/0.0739) = 27.4 m/s

Using the information provided about the power, solve for the angular frequency omega (ù). Divide that by 2 pi for the frequency in Hz.

Answer 2

My symbols for omega and mu did not come out in Greek like they should, but I hope you get the idea.