A pipe closed at one end has length $0.8 \mathrm{~m}$. At its open end $0.5 \mathrm{~m}$ long uniform string…
A pipe closed at one end has length $0.8 \mathrm{~m}$. At its open end $0.5 \mathrm{~m}$ long uniform string is vibrating in its $2^{\text {nd }}$ harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is $50 \mathrm{~N}$ and the speed of sound is $320 \mathrm{~m} / \mathrm{s}$, the mass of the string is
20 gram
10 gram
5 gram
15 gram
Solution
$\begin{aligned} & 2 \times\left[\frac{1}{2 \ell_1} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}\right]=\frac{v}{4 \ell_2} \\ & \frac{1}{0.5} \sqrt{\frac{50}{\mathrm{~m}}}=\frac{320}{4 \times 0.8} \\ & \therefore \mathrm{m}=0.02 \mathrm{~kg} / \mathrm{m} \\ & \therefore \text { Total mass of the string } \\ & =0.02 \times 0.5 \mathrm{~kg}=10 \mathrm{gm}\end{aligned}$