Physics - Sound Wave Question with Solution | TestHub

For a closed organ pipe, the fundamental frequency is given $f=\frac{v}{4L}$, where $v=320\mathrm{m}/\mathrm{s}$ is the velocity of sound in the medium of organ pipe and $L=0.8\mathrm{m}$ being the length of pipe. Now, we are given second harmonic frequency of wire is equal and in resonance with fundamental frequency of pipe, thus; $f=\frac{v}{4L}=\frac{1}{l}\sqrt{\frac{T}{\mu }}$ Putting, $T=50\mathrm{N}$ and $l=0.8\mathrm{m}$ $\begin{array}{rl}& \frac{320}{4(0.8)}=\frac{1}{0.5}\sqrt{\frac{50}{\mu }}\\ & \Rightarrow \sqrt{\frac{\mu }{50}}=\frac{1}{50}\end{array}$ The length of string $=0.5\mathrm{m}$ Thus, mass of string $=0.02\times 0.5=0.01\mathrm{kg}=10\mathrm{g}$