Question

A $25cm$ long string is fixed at both ends. It has a mass of $2.5g$ and vibrates in its first overtone under tension. Also given is a closed pipe $30cm$ 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 $360m{s}^{-1}$ )

Answer 1

Answer: 5929

$\mu =\frac{2.5}{25}=0.1g/cm=10^{-2}kg/m$
$1^{\text{st }}$ overtone,
${\lambda }_s=25cm=0.25m$
$\Rightarrow f_s=\frac{1}{{\lambda }_s}\sqrt{\frac{T}{\mu }}$

Pipe in fundamental frequency, $\frac{{\lambda }_p}{4}=0.3$
${\lambda }_p=1.2m$
$\Rightarrow f_p=\frac{v_s}{{\lambda }_p}=\frac{360}{1.2}=300Hz$

$\because$ By decreasing the tension, beat
frequency is decreased.
$\therefore f_s>f_p$
$\Rightarrow f_s-f_p=8$
$\Rightarrow \frac{1}{0.25}\sqrt{\frac{T}{10^{-2}}}-\frac{360}{1.2}=8$
$\Rightarrow \frac{10\times 100}{25}\sqrt{T}-300=8$
$\Rightarrow \sqrt{T}=\frac{308\times 25}{1000}$
$\Rightarrow T=59.29N$
$\Rightarrow 100T=5929N$