A string fixed at both ends is in resonance i | vedclass

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Question

(C) For a string fixed at both ends,the frequency of the $2^{nd}$ harmonic is given by $f_1 = \frac{2v}{2L} = \frac{v}{L}$,where $v$ is the wave speed

Vedclass · Accepted Answer

$n = 5, f_2 = \frac{5}{4}f_1$

Vedclass · Answer

(C) For a string fixed at both ends,the frequency of the $2^{nd}$ harmonic is given by $f_1 = \frac{2v}{2L} = \frac{v}{L}$,where $v$ is the wave speed and $L$ is the length of the string.When one end becomes free,the string acts like a pipe closed at one end. The resonant frequencies are given by $f_n = \frac{nv}{4L}$,where $n$ must be an odd integer $(n = 1, 3, 5, \dots)$.We are given that the frequency is increased from $f_1$. We need to find the first resonant frequency $f_2 > f_1$ for the new configuration.Setting $f_2 = \frac{nv}{4L} > f_1 = \frac{v}{L}$,we get $\frac{nv}{4L} > \frac{v}{L}$,which implies $n > 4$.The smallest odd integer greater than $4$ is $n = 5$.Thus,$f_2 = \frac{5v}{4L} = \frac{5}{4} \left( \frac{v}{L} \right) = \frac{5}{4}f_1$.

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