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(C) Given: Linear mass density $\mu = 0.1 \ kg/m$,Length $L = 0.9 \ m$,Tension $T = 40 \ N$,Amplitude $A = 0.3 \ cm = 0.003 \ m$,Number of segments $n

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$\frac{\pi}{5}$

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(C) Given: Linear mass density $\mu = 0.1 \ kg/m$,Length $L = 0.9 \ m$,Tension $T = 40 \ N$,Amplitude $A = 0.3 \ cm = 0.003 \ m$,Number of segments $n = 3$.Wave speed $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{40}{0.1}} = \sqrt{400} = 20 \ m/s$.The frequency of the $n^{th}$ harmonic is $f_n = \frac{n v}{2L} = \frac{3 	imes 20}{2 	imes 0.9} = \frac{60}{1.8} = \frac{600}{18} = \frac{100}{3} \ Hz$.Angular frequency $\omega = 2 \pi f_n = 2 \pi 	imes \frac{100}{3} = \frac{200 \pi}{3} \ rad/s$.Maximum particle velocity $v_{max} = A \omega = 0.003 	imes \frac{200 \pi}{3} = \frac{3}{1000} 	imes \frac{200 \pi}{3} = \frac{\pi}{5} \ m/s$.

$A$ string of mass $0.1 \ kg \ m^{-1}$ has length $0.9 \ m$. It is fixed at both ends and stretched such that it has a tension of $40 \ N$. The string vibrates in three segments with amplitude $0.3 \ cm$. The amplitude (maximum) of the particle velocity is (in $m/s$):

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