Question

A $25\,cm$ long string is fixed at both ends. It has a mass of $2.5\,g$ and vibrates in its first overtone under tension. Also given is a closed pipe $30\,cm$ long which vibrates in its fundamental frequency. When both of these are sounded simultaneously $8$ beats per second are heard. It is also observed that decreasing the tension in the string also decreases the beat frequency. What is the value of tension $\left(\left(10\right)^{2} T\right)$ in the string (in $N$ )? (Given that the speed of sound in air is $360\,ms^{- 1}$ )

Answer 1

$\mu =\frac{2 . 5}{25}=0.1\,g/cm=10^{- 2}kg/m$
$1^{\text{st }}$ overtone,
$\lambda _{s}=25\,cm=0.25\,m$
$\Rightarrow f_{s}=\frac{1}{\lambda _{s}}\sqrt{\frac{T}{\mu }}$

Pipe in fundamental frequency, $\frac{\lambda _{p}}{4}=0.3$
$\lambda _{p}=1.2\,m$
$\Rightarrow f_{p}=\frac{v_{s}}{\lambda _{p}}=\frac{360}{1 . 2}=300\,Hz$

$\because $ By decreasing the tension, beat
frequency is decreased.
$\therefore f_{s}>f_{p}$
$\Rightarrow f_{s}-f_{p}=8$
$\Rightarrow \frac{1}{0 . 25}\sqrt{\frac{T}{10^{- 2}}}-\frac{360}{1 . 2}=8$
$\Rightarrow \frac{10 \times 100}{25}\sqrt{T}-300=8$
$\Rightarrow \sqrt{T}=\frac{308 \times 25}{1000}$
$\Rightarrow T=59.29\,N$
$\Rightarrow 100\,T=5929\,N$