Question

↵A hollow pipe of length 0.08m is is closed at one end. at its open end a 0.5 m long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. if the tension in the wire is 50 N and the speed of sound is 320 ms^-1, the mass of the string is

1. 5 g
2. 10 g
3. 20g
4. ​40 g

Answer 1

$$\begin{gathered} \mathrm{Fundamental}\mathrm{frequency}\mathrm{of}\mathrm{closed}\mathrm{pipe}: \\ \mathrm{f}=\frac{\mathrm{v}}{4\mathrm{L}} \\ =\frac{320}{4\times 0.08} \\ =1000\mathrm{Hz} \\ \mathrm{Second}\mathrm{harmonic}\mathrm{of}\mathrm{the}\mathrm{string}: \\ {\mathrm{n}}_2=2\mathrm{n} \\ =2\left(\frac{1}{2\mathrm{L}}\right)\sqrt{\frac{\mathrm{T}}{\mathrm{m}/\mathrm{L}}} \\ =\frac{1}{0.5}\sqrt{\frac{50\times 0.5}{\mathrm{m}}} \\ =\frac{10}{\sqrt{\mathrm{m}}} \\ \mathrm{But},{\mathrm{n}}_2=\mathrm{f}=1000\mathrm{Hz} \\ \mathrm{Therefore}, \\ 1000=\frac{10}{\sqrt{\mathrm{m}}} \\ \sqrt{\mathrm{m}}=\frac{10}{1000}\mathrm{kg} \\ =10\mathrm{g} \\ \mathbf{Answer}\mathbf{:}\mathbf{}\mathbf{Option}\mathbf{}\mathbf{-}\mathbf{}\mathbf{\left(}\mathbf{2}\mathbf{\right)} \end{gathered}$$