Question

An organ pipe $P_1$ closed at one end and vibrating in its first overtone pipe $P_2$ open at both ends vibrating in its third overtone are in resonance with a given tuning fork. The ratio of the lengths of $P_1$ to that of $P_2$ is?

Answer 1

Step 1: Given data

1. Organ pipe $P_1$ closed at one end and vibrating in its first overtone pipe
2. Pipe $P_2$ open at both ends vibrating in its third overtone
3. The vibration of both ends is in resonance

Step 2: Formula used

1. $v=n\lambda$ where $v$ is the velocity of the wave, $n$ is the frequency of the wave and $\lambda$ is the wavelength of the wave.

Step 3: Find the frequency at $P_1$ and $P_2$

Let $L_1$ and $L_2$ be the length of $P_1$ and $P_2$

The first overtone of the closed organ pipe is $=\frac{3\lambda }{4}=L_1$

$$\begin{gathered} \Rightarrow \lambda =\frac{4L_1}{3} \\ \therefore n_1=\frac{v}{{\displaystyle \frac{4L_1}{3}}} \\ =\frac{3V}{4L_1} \end{gathered}$$

The third overtone of an open organ pipe is $=\frac{4\lambda }{2}=L_2$

$$\begin{gathered} \Rightarrow \lambda =\frac{L_2}{2} \\ \therefore n_2=\frac{v}{{\displaystyle \frac{L_2}{2}}} \\ =\frac{2V}{L_2} \end{gathered}$$

Step 4: Find the ratio of the lengths of $P_1$ to that of $P_2$

Let $n_1$ is the frequency for $P_1$ and $n_2$ is the frequency for $P_2$

Given that organ, pipe $P_1$ closed at one end and vibrating in its first overtone pipe $P_2$ open at both ends and vibrating in its third overtone are in resonance with a given tuning fork

$$\begin{gathered} \therefore n_1=n_2 \\ \Rightarrow \frac{3V}{4L_1}=\frac{2V}{L_2} \\ \Rightarrow \frac{L_1}{L_2}=\frac{3}{8} \end{gathered}$$

Hence the ratio of the lengths of $P_1$ to that of $P_2$ is $3:8$