Question

A horizontal wire is stretched with a tension of 98.4 N, and the speed of transverse waves for the wire is 406 m/s. What must the amplitude of a traveling wave of frequency 60.5 Hz be for the average power carried by the wave to be 0.391 W? Give an answer in mm. Pay attention to the number of significant figures.

Answer 1

Answer:

Amplitude = 4.725 mm

Explanation:

We have got the following values :

T = 98.4 N

T is the wire tension

v = 406 m/s

v is the transverse waves speed

f = 60.5 Hz

f is the frequency

ΔU = 0.391 W

ΔU is the average power carried by the wave

If μ is  the mass per unit length of the wire ⇒ μ = $\frac{T}{v^{2}}$

If ω is the angular frequency of the wire ⇒ ω = 2π.f

So ω = 2π(60.5 Hz) and  ω units are $s^{-1}$

The equation that relates all is :

ΔU = (1/2).(T/v^2).(ω^2).(A^2).v

Where A is the wave amplitude

$0.391 W=\frac{1}{2}.\frac{98.4 N}{(406\frac{m}{s})^{2}}.(2.pi.(60.5Hz))^{2}.A^{2}.406\frac{m}{s}$

This will give us A in meters

$A=4.7253(10^{-3})m\\A=4.725 mm$