A stretched uniform wire of length L under te | vedclass

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Question

(B) The fundamental frequency of a stretched wire is $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.The fundamental frequency of a closed pipe of length $L$

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$2$

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(B) The fundamental frequency of a stretched wire is $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.The fundamental frequency of a closed pipe of length $L$ is $n = \frac{v}{4L}$.Given that both have the same initial frequency, we have $\frac{1}{2L} \sqrt{\frac{T}{\mu}} = \frac{v}{4L}$.When tension is increased by $16 \,N$, the new frequency $n'$ is $n' = \frac{1}{2L} \sqrt{\frac{T+16}{\mu}}$.This new frequency resonates with the $2^{	ext{nd}}$ harmonic of the closed pipe. The harmonics of a closed pipe are odd multiples of the fundamental frequency $(n, 3n, 5n, \dots)$. However, the question specifies the $2^{	ext{nd}}$ harmonic, which implies the $3n$ frequency (as the $1^{	ext{st}}$ is $n$, $2^{	ext{nd}}$ is $3n$).Thus, $n' = 3n$.$\frac{1}{2L} \sqrt{\frac{T+16}{\mu}} = 3 \left( \frac{1}{2L} \sqrt{\frac{T}{\mu}} ight)$.Squaring both sides: $\frac{T+16}{T} = 3^2 = 9$.$T+16 = 9T \implies 8T = 16 \implies T = 2 \,N$.

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