Question

A string $25cm$ long, having a mass $2.5gm$ is under tension. A pipe closed at one end is $40cm$ long. When the string is set vibrating in its first overtone and the air column in the pipe is in its fundamental frequency, $8$ beats are heard per second. It is observed that decreasing the tension in the string decreases the beat frequency. If the speed of sound in air is $320m{s}^{-1}$, the tension in the string is nearly

• A. $27N$
• B. $54N$
• C. $13.5N$
• D. $108N$

Answer 1

Answer: (A)

$27N$

For a closed pipe, the fundamental frequency is given as: ${\nu }_1=\frac{v}{4L}$

where $v=320m{s}^{-1},L=0.4m$

So, ${\nu }_1=200Hz$

Let the fist overtone or second harmonic frequency of standing wave on string be ${\nu }_2$.

So, ${\nu }_2=\frac{\sqrt{T/\mu }}{\left(2L/2\right)}$

where $T$ is the tension in the string, $L=0.25m$ is the length of the string, and $\mu =0.0025/0.25=0.01kg/m$ is the linear mass density.

Now beats frequency is ${\nu }_1\pm {\nu }_2=8Hz\Rightarrow {\nu }_2=200\pm 8=192Hzor208Hz$

As decreasing tension decreases, the frequency of string decreases and beats frequency decreases. So frequency of string should be greater than that of pipe.

So, ${\nu }_2=208Hz$

Now, $208=\frac{\sqrt{T/0.01}}{0.25}\Rightarrow T=(208\times 0.25{)}^2\times 0.01=27.04N$