Question

What are (a) the lowest frequency, (b) the second lowest frequency, and (c) the third lowest frequency for standing waves on a wire that is 10.9 m long, has a mass of 55.8 g, and is stretched under a tension of 253 n?

Answer 1

(a) The lowest frequency (called fundamental frequency) of a wire stretched under a tension T is given by

`f_1 = (1)/(2L) \sqrt{ (T)/(m/L) }`
where
L is the wire length
T is the tension
m is the wire mass

In our problem, L=10.9 m, m=55.8 g=0.0558 kg and T=253 N, therefore the fundamental frequency of the wire is

`T= (1)/(2L) \sqrt{ (T)/(m/L) }= (1)/(2 \cdot 10.9 m) \sqrt{ (253 N)/(0.0558 kg/10.9 m) }= 10.2 Hz`

b) The frequency of the nth-harmonic for a standing wave in a wire is given by

`f_n = n f_1`
where n is the order of the harmonic and f1 is the fundamental frequency. If we use n=2, we find the second lowest frequency of the wire:

`f_2 = 2 f_1 = 2 \cdot 10.2 Hz=20.4 Hz`

c) Similarly, the third lowest frequency (third harmonic) is given by

`f_3 = 3 f_1 = 3 \cdot 10.2 Hz = 30.6 Hz`