The fundamental frequency of a sonometer wire | vedclass

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(B) The fundamental frequency of a sonometer wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.Since $T$ and $\mu$ are constant,$f \propto \fra

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decreased by $20 \%$

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(B) The fundamental frequency of a sonometer wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.Since $T$ and $\mu$ are constant,$f \propto \frac{1}{L}$.Let the initial length be $L_1 = L$ and the initial fundamental frequency be $f_1 = 50 	ext{ Hz}$.The new length is $L_2 = L + 0.25L = 1.25L = \frac{5}{4}L$.The new fundamental frequency $f_2$ is given by $\frac{f_2}{f_1} = \frac{L_1}{L_2} = \frac{L}{1.25L} = \frac{1}{1.25} = 0.8$.So,$f_2 = 0.8 	imes 50 	ext{ Hz} = 40 	ext{ Hz}$.The frequency of the second harmonic is $f'_n = n 	imes f_n$. For the second harmonic $(n=2)$,$f'_2 = 2 	imes f_2 = 2 	imes 40 	ext{ Hz} = 80 	ext{ Hz}$.The initial frequency of the second harmonic was $f'_1 = 2 	imes f_1 = 2 	imes 50 	ext{ Hz} = 100 	ext{ Hz}$.The change in frequency is $\Delta f = f'_2 - f'_1 = 80 - 100 = -20 	ext{ Hz}$.The percentage change is $\frac{\Delta f}{f'_1} 	imes 100 = \frac{-20}{100} 	imes 100 = -20 \%$.Thus,the frequency of the second harmonic decreases by $20 \%$.

The fundamental frequency of a sonometer wire is $50 \text{ Hz}$ for a given length and tension. If the length is increased by $25 \%$ while keeping the tension constant,what is the percentage change in the frequency of the second harmonic?

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