(a) The lowest frequency (called fundamental frequency) of a wire stretched under a tension T is given by
[tex]f_1 = \frac{1}{2L} \sqrt{ \frac{T}{m/L} } [/tex]
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
[tex]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[/tex]
b) The frequency of the nth-harmonic for a standing wave in a wire is given by
[tex]f_n = n f_1[/tex]
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:
[tex]f_2 = 2 f_1 = 2 \cdot 10.2 Hz=20.4 Hz[/tex]
c) Similarly, the third lowest frequency (third harmonic) is given by
[tex]f_3 = 3 f_1 = 3 \cdot 10.2 Hz = 30.6 Hz[/tex]
