Question

Two wires, each of length 3.15 m, are stretched between two fixed supports. On wire A, there is a second-harmonic standing wave whose frequency is 878.0 Hz. However, the same frequency of 878.0 Hz is the third harmonic on wire B. Find the speed at which the individual waves travel on each wire and give the speed difference (m/s) of the two waves as the final answer.

Answer 1

To find the wave speeds on two wires vibrating at different harmonics but at the same frequency, we use the standing wave formula. This enables the calculation of individual wave speeds and the determination of the speed difference.

Step-by-step explanation:

The question involves finding the wave speed on two different wires and the difference in speed between them, given the harmonic frequencies. For wire A, which exhibits the second harmonic at 878.0 Hz, we use the formula for standing waves on a string: frequency (f) = n * (v / 2L), where n is the harmonic number, v is the wave speed, and L is the length of the wire. Similarly, for wire B, which exhibits the third harmonic, the same formula applies with n being 3.

For wire A, with L = 3.15 m and n = 2, the equation becomes 878.0 Hz = 2 * (v / 2 * 3.15), which gives a wave speed v = 878.0 Hz * 2 * 3.15. For wire B, with the same L and frequency but n = 3, the speed of the wave is calculated similarly. Lastly, the speed difference is found by subtracting the speeds of wire A and wire B.