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
$f_1 = \frac{1}{2L} \sqrt{ \frac{T}{m/L} }$
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
$T= \frac{1}{2L} \sqrt{ \frac{T}{m/L} }= \frac{1}{2 \cdot 10.9 m} \sqrt{ \frac{253 N}{0.0558 kg/10.9 m} }= 10.2 Hz$

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