Question

A horizontal wire is stretched with a tension of 94.0 N , and the speed of transverse waves for the wire is 406 m/s.

What must the amplitude of a traveling wave of frequency 73.0 Hz be in order for the average power carried by the wave to be 0.360 W ?

Answer 1

Answer 2

To find the amplitude of the traveling wave, we can use the formula for the average power carried by a wave:

P_avg = (1/2) * ρ * v * ω^2 * A^2

where P_avg is the average power, ρ is the linear mass density of the wire, v is the speed of the wave, ω is the angular frequency, and A is the amplitude of the wave.

In this case, we're given: P_avg = 0.360 W v = 406 m/s f = 73.0 Hz

The angular frequency ω can be calculated from the frequency f using the formula: ω = 2πf

Now, we need to find the linear mass density ρ of the wire. We can use the tension T and the speed of the wave v to find ρ using the formula: v = √(T/ρ)

Solving for ρ, we have: ρ = T / v^2

Substituting the given values, we find: ρ = (94.0 N) / (406 m/s)^2

Now, we can calculate the angular frequency ω: ω = 2π * (73.0 Hz)

Next, we can rearrange the formula for average power and solve for the amplitude A: A = √(2 * P_avg / (ρ * v * ω^2))

Substituting the known values, we can calculate A:

A = √(2 * 0.360 W / ((94.0 N) / (406 m/s)^2 * 406 m/s * (2π * 73.0 Hz)^2))

Simplifying the equation, we get:

A ≈ 0.00182 m

Therefore, the amplitude of the traveling wave should be approximately 0.00182 m.