The fundamental frequency of sonometer wire is 50 Hz for some length and tension. If the length is increased by 25% by keeping the tension same, then the frequency change of second harmonic is

Question

TestHub · Accepted Answer

The fundamental frequency of a sonometer wire is given by $f = \frac{1}{2 L} \frac{T}{μ}$ .

Since tension $T$ and mass per unit length $μ$ are constant, the frequency is inversely proportional to the length, $f \propto \frac{1}{L}$ .

The frequency of the second harmonic is $f_{2} = 2 f$ , which also follows $f_{2} \propto \frac{1}{L}$ .

When the length is increased by $25%$ , the new length becomes $L^{'} = L + 0.25 L = 1.25 L = \frac{5}{4} L$ .

The new frequency of the second harmonic is $f_{2}^{'} = f_{2} \times \frac{L}{L ^{'}} = f_{2} \times \frac{4}{5} = 0.8 f_{2}$ .

The percentage change in the frequency of the second harmonic is $\frac{f _{2}^{'} - f _{2}}{f _{2}} \times 100% = \frac{0.8 f _{2} - f _{2}}{f _{2}} \times 100% = - 20%$ .

Thus, the frequency of the second harmonic decreases by $20%$ .

Answer: decrease by $20%$

Physics - Wave on String Question with Solution | TestHub

Options:

Doubts & Discussion