Question

An open pipe is suddenly closed one end with the result that the frequency of the third harmonic of the closed pipe is found to be higher by $100Hz$ then the fundamental frequency of the open pipe.
The fundamental frequency of the open pipe is:

• A. $200Hz$
• B. $300Hz$
• C. $240Hz$
• D. $480Hz$

Answer 1

The correct option is A $200Hz$

For proton to move at ${45}^o$

$qE=mg$

so that the net force is along ${45}^o$

Now $qE=mg$ and $E=\frac{V}{d}$

$\frac{1.6}{{10}^{19}}\times \frac{V\times 100}{1}=\frac{1.6}{{10}^{27}}\times 10$

$V=1\times {10}^{-9}V$For closed pipe.

$f=\frac{(2n+1)v}{4l}$

$n=1$ for third harmonic

$f_1=\frac{3}{4}\frac{v}{l}$

for open pipe $f=\frac{nv}{2l}.$

$f_2=\frac{1\times v}{2l}=\frac{v}{2l}$

$\therefore x=1$

$ATQ$

$f_1-f_2=100$

$\Rightarrow \frac{3}{4}\frac{v}{l}-\frac{v}{2l}=100$

$\Rightarrow \frac{v}{l}\times \frac{1}{4}=100$

$\Rightarrow \frac{v}{l}=400$

$f_2=\frac{v}{2l}=\frac{1}{2}\times 400=200H_2$