Question

Two wires, each of length 1.3 m, are stretched between two fixed supports. On wire A there is a second-harmonic standing wave whose frequency is 640 Hz. However, the same frequency of 640 Hz is the third harmonic on wire B. Find the speed at which the individual waves travel on each wire.

Answer 1

Answer:

Explanation:

Given

Length of each wire $L=1.3\ m$

On wire A second harmonic frequency is given by

$f_2_{a}=2\times (\frac{v}{2L})$

where f=frequency

v=velocity of wave

L=length of wire

$v_a=f_2\times L$

$v_a=640\times 1.3=832\ m/s$

For wire B third harmonic is given by

$f_3_{b}=3\times (\frac{v}{2L})$

$v_b=\frac{2L}{3}\cdot f_3_{b}$

$v_b=\frac{2\times 1.3}{3}\times 640=554.66\ m/s$