Question

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

• A. 5
• B. 10
• C. 20
• D. 40

Answer 1

The correct option is B 10

The fundamental mode in a pipe closed at one end and the second harmonic in a string are shown in figure. it can be seen that ${\lambda }_p/4=L_{p}and{\lambda }_s=L_s$

For the pipe closed at one end.

$v_p=\frac{v_p}{{\lambda }_p}=\frac{v_p}{4L_p}=\frac{320}{4\left(0.8\right)}=100Hz.$

where $v_p=320m/s$ is the velocity of sound in the pipe and $L_p=0.8m$ is length of the pipe. For string of mass $m$, length $L_s$ and having tension $T$, velocity of sound in the string is

given by,

$v_p=\frac{v_p}{=}\frac{\sqrt{T/\left(m/L_s\right)}}{l_s}=\sqrt{\frac{T}{m{L}_s}}=\sqrt{\frac{50}{m\left(0.5\right)}}=\frac{10}{\sqrt{m}}$

At resonance, $v_p=v_s.$Subsitute $v_p$ and $v_s$ from first and second equations to get $m=0.01kg=10gram$