A hollow pipe of length $0.8 \mathrm{~m}$ is closed at one end. At its open end, a $0.8 \mathrm{~m}$ long…
A hollow pipe of length $0.8 \mathrm{~m}$ is closed at one end. At its open end, a $0.8 \mathrm{~m}$ long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of pipe. If the tension in the sting is $50 \mathrm{~N}$ and speed of sound in air is $320 \mathrm{~m} / \mathrm{s}$, the mass of the strings is
$10 \mathrm{~g}$
$20 \mathrm{~g}$
$5 \mathrm{~g}$
$40 \mathrm{~g}$
Solution
For a closed organ pipe, the fundamental frequency is given $f=\frac{v}{4 L}$, where $v=320 \mathrm{~m} / \mathrm{s}$ is the velocity of sound in the medium of organ pipe and $L=0.8 \mathrm{~m}$ being the length of pipe.
Now, we are given second harmonic frequency of wire is equal and in resonance with fundamental frequency of pipe, thus;
$f=\frac{v}{4 L}=\frac{1}{l} \sqrt{\frac{T}{\mu}}$
Putting, $T=50 \mathrm{~N}$ and $l=0.8 \mathrm{~m}$
$\begin{aligned} & \frac{320}{4(0.8)}=\frac{1}{0.5} \sqrt{\frac{50}{\mu}} \\ & \Rightarrow \sqrt{\frac{\mu}{50}}=\frac{1}{50}\end{aligned}$
The length of string $=0.5 \mathrm{~m}$
Thus, mass of string $=0.02 \times 0.5=0.01 \mathrm{~kg}=10 \mathrm{~g}$