The closed and open organ pipe have same length and when they are vibrating simultaneously in first overtone produce 3 beats. The length of open pipe is made ( 3 1 ) rd and that of closed pipe is ma

Question

The closed and open organ pipe have same length and when they are vibrating simultaneously in first overtone produce 3 beats. The length of open pipe is made ( 3 1 ​ ) rd and that of closed pipe is made 3 times the original, the number of beats produced in fundamental mode will be (Neglect end correction)

TestHub · Accepted Answer

Let the original length of both the open and closed organ pipes be $L$ and the speed of sound be $v$ .

The frequency of the first overtone of the open pipe is $n_{1} = \frac{v}{L}$ .

The frequency of the first overtone of the closed pipe is $n_{2} = \frac{3 v}{4 L}$ .

Given that they produce 3 beats per second in their first overtone:

$n_{1} - n_{2} = 3$

$\frac{v}{L} - \frac{3 v}{4 L} = 3$

$\frac{v}{4 L} = 3 \Rightarrow \frac{v}{L} = 12$

Now, the length of the open pipe is made $L_{1} = \frac{L}{3}$ .

The fundamental frequency of the new open pipe is:

$n_{1}^{'} = \frac{v}{2 L _{1}} = \frac{v}{2 ( \frac{L}{3} )} = \frac{3 v}{2 L} = \frac{3}{2} \times 12 = 18 Hz$

The length of the closed pipe is made $L_{2} = 3 L$ .

The fundamental frequency of the new closed pipe is:

$n_{2}^{'} = \frac{v}{4 L _{2}} = \frac{v}{4 ( 3 L )} = \frac{v}{12 L} = \frac{1}{12} \times 12 = 1 Hz$

The number of beats produced in the fundamental mode is:

$n_{1}^{'} - n_{2}^{'} = 18 - 1 = 17$

Answer: $17$

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