Question

A hollow pipe of length $0.8\text{m}$ is closed at one end. At its open end a $0.5\text{m}$ long uniform string (fixed at both ends) is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is $50\text{N}$ and the speed of sound is $320{\text{ms}}^{-1}$, the mass of the string is

• A. $5\text{g}$
• B. $10\text{g}$
• C. $20\text{g}$
• D. $40\text{g}$

Answer 1

The correct option is B $10\text{g}$

For resonace, the frequency of vibration of hollow pipe must be equal to the frequency of vibration of long string (fixed both ends).


Given fundamental frequency of $I$ = second harmonic of $II$

$\Rightarrow \frac{v_2}{4l_2}=2\frac{v_1}{2l_1}.....\left(1\right)$

Here, $v_2=320\text{m/s},l_2=0.8\text{m},$

$v_1=\sqrt{\frac{T}{\mu }},l_1=0.5\text{m}$

Using these in $\left(1\right)$

$\Rightarrow \frac{320}{4\times 0.8}=\frac{1}{0.5}\sqrt{\frac{T}{\mu }}$

or, $\frac{(320{)}^2}{16\times (0.8{)}^2}=(2{)}^2\times \frac{T}{\mu }$

or, $\frac{(320{)}^2}{16\times (0.8{)}^2}=\frac{4\times 50}{\mu }$

$[T=50\text{N}]$

or, $\mu =\frac{16\times 4\times (0.8{)}^2\times 50}{(320{)}^2}$

$\therefore \mu =0.02\text{kg/m}$

Thus mass of string is,

$m_1=\mu l_1=0.02\times 0.5=0.01\text{kg}$

$\Rightarrow m_1=0.01\times 1000=10\text{g}$