A pipe open at one end has length $0.8 \mathrm{~m}$. At the open end of the tube a string $0.5 \mathrm{~m}$…
A pipe open at one end has length $0.8 \mathrm{~m}$. At the open end of the tube a string $0.5 \mathrm{~m}$ long is vibrating in its $1^{\text {st }}$ overtone and resonates with fundamental frequency of pipe. If tension in the string is $50 \mathrm{~N}$, the mass of string is (speed of sound $=320 \mathrm{~m} / \mathrm{s}$)
25 gram
15 gram
20 gram
10 gram
Solution
$2 \times\left[\frac{1}{2 \ell_{1}} \sqrt{\frac{T}{m}}\right]=\frac{v}{4 \ell_{2}}$
$\frac{1}{0.5} \sqrt{\frac{50}{m}}=\frac{320}{4 \times 0.8}$
$\therefore \quad$ Total mass of the string $=0.02 \times 0.5 \mathrm{~kg}=10 \mathrm{gm}$