Question

A string $25\text{cm}$ long and having a mass of $2.5\text{gm}$ is under tension. A pipe closed at one end is $40\text{cm}$ long. When the string is set vibrating in its first overtone and the air in the pipe in its fundamental frequency, $8$ beats per second are heard. It is observed that decreasing the tension in the string decreases beat frequency. If the speed of sound in air is $320\text{m/s}$, the tension in the string is N.

Answer 1

Frequency of fundamental mode of closed pipe,
$f_1=\frac{v}{4l}=200Hz.$
Decreasing the tension in the string decrease the beat frequency. Hence, the first overtone frequency of the string should be $208Hz$ (not $192Hz)$
$208=\frac{1}{l}\sqrt{\frac{T}{\mu }}$
$T=\mu (l208{)}^2$
$T=(2.5\times \frac{{10}^{-3}}{0.25})\left(0.25{)}^2\right(208{)}^2$
$T=27.04$ N