A pipe closed at one end has lengh $0 \cdot 8 \mathrm{~cm}$. At its open end a $0.5 \mathrm{~m}$ long…
A pipe closed at one end has lengh $0 \cdot 8 \mathrm{~cm}$. At its open end a $0.5 \mathrm{~m}$ long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is $50 \mathrm{~N}$ and the speed of sound is $320 \mathrm{~m} / \mathrm{s}$, the mass of the string is
8 gram
2 gram
10 gram
4 gram
Solution
$\begin{aligned}
& \ell=\text { length of string }=0.5 \mathrm{~m} \\
& \mathrm{~L}=\text { length of pipe }=0.8 \mathrm{~m} \\
& \text { frequency of string vibrating in the second harmonics is } \\
& \mathrm{V}=(2 / 2 \ell) \mathrm{V}(\mathrm{~T} / \mu) \text { Here } \mu \text { is mass per unit length of string. }
\end{aligned}$
Fundamental frequency of closed pipe is
$V^{\prime}=\{U / 4 L\}$ where $U$ is speed of sound
For resonance $\mathrm{V}=\mathrm{V}$ ' hence
$(2 / 2 \ell) \sqrt{ }(T / \mu)=(U / 4 L)$
$\therefore\{1 /(0.5)\} \vee(50 / \mu)=\{(320) /(4 \times 0.8)\}$
$\therefore \sqrt{ }(50 / \mu)=\{(160) /(4 \times 0.8)\}$
$\therefore(50 / \mu)=2500$
$\therefore \mu=\{50 /(2500)\}=(1 / 50)$
$\therefore$ mass of string $=m=\mu \ell=(1 / 50) \times 0.5=\{1 /(100)\} \mathrm{kg}=10 \mathrm{gm}$