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 = [tex]\frac{T}{v^{2}}[/tex]
If ω is the angular frequency of the wire ⇒ ω = 2π.f
So ω = 2π(60.5 Hz) and ω units are [tex]s^{-1}[/tex]
The equation that relates all is :
ΔU = (1/2).(T/v^2).(ω^2).(A^2).v
Where A is the wave amplitude
[tex]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}[/tex]
This will give us A in meters
[tex]A=4.7253(10^{-3})m\\A=4.725 mm[/tex]
