Question

A hollow pipe of length $0.8m$ is closed at one end. At its open end a $0.5m$ 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 $50N$ and the speed of sound is $320m{s}^{-1},$ the mass of the string is

• A. 5 g
• B. 10 g
• C. 20 g
• D. 40 g

Answer 1

Answer: (B)

Given, $2nd$ harmonic of $I=$ Fundamental of $II$
$\therefore 2\left(\frac{v_1}{2l_1}\right)=\frac{v_2}{4l_2}$
$\Rightarrow \frac{T/\mu }{l_1}=\frac{v_2}{4l_2}$
$\Rightarrow \mu =\frac{16T{l}_2^2}{v_2^2{l}_1^2}$
$=\frac{16\times 50\times (0.8{)}^2}{(320{)}^2\times (0.5{)}^2}$
$=0.02kg/m$
$\therefore m_1=\mu l_1=(0.02)(0.2)$
$=0.01kg=10g$