A pipe closed at one end length $0.8 \mathrm{~m}$. At its open end a $0.5 \mathrm{mlong}$ uniform string is…
A pipe closed at one end length $0.8 \mathrm{~m}$. At its open end a $0.5 \mathrm{mlong}$ uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire in $50 \mathrm{~N}$ and the speed of sound is $320 \mathrm{~m} / \mathrm{s}$, the mass of the string used is
$8 \mathrm{~g}$
$4 \mathrm{~g}$
$12 \mathrm{~g}$
$10 \mathrm{~g}$
Solution
For fundamental harmonics of a closed end pipe, $\frac{\lambda}{4}=L$
Velocity of sound is in air is given by, $v=\lambda f$
For second harmonics of a string, $\lambda^{\prime}=L^{\prime}$ the wavelength is the length of the string.
Speed of sound on the wave is related as, $v_{\mathrm{S}}=\lambda^{\prime} f$.
Speed of sound is given by, $v_{\mathrm{S}}=\sqrt{\frac{T}{\mu}}, T$ is the tension in the wire and $\mu$ the mass per unit length of the string.
Since, the string in second harmonics is in resonance with the closed end pipe in fundamental mode,
$\begin{aligned} & f=\frac{v}{4 L}=\frac{v_{\mathrm{S}}}{L^{\prime}}=\frac{1}{L^{\prime}} \sqrt{\frac{T}{\mu}} \\ & \Rightarrow \frac{320 \mathrm{~m} / \mathrm{s}}{4 \times 0.8 \mathrm{~m}}=\frac{1}{0.5 \mathrm{~m}} \sqrt{\frac{50 \mathrm{~N}}{\mu}} \\ & \Rightarrow \mu=\frac{1}{50} \mathrm{kgm}^{-1} \\ & \Rightarrow m=\mu L^{\prime}=\frac{0.5 \mathrm{~m}}{50 \mathrm{~kg}^{-1} \mathrm{~m}}=10 \mathrm{~g}\end{aligned}$
^