Question

A pipe closed at one end produces a fundamental note of $412Hz$. It is cut into two pieces of equal length. The fundamental frequencies produced in the two pieces are

• A. $206Hz,412Hz$
• B. $824Hz,1648Hz$
• C. $412Hz,824Hz$
• D. $206Hz,824Hz$

Answer 1

The correct option is B

$824Hz,1648Hz$

Explanation for the correct answer is:

Step 1. Given Data:

Length of the pipe initially be $L$.
Fundamental frequency in closed pipe, $$\begin{gathered} f=\frac{V}{4L}\dots \left(1\right) \\ \Rightarrow f=412Hz \end{gathered}$$
After cutting, one pipe becomes open at both ends and the other remains as closed at one end type.
Let the Length of closed at one end and open at both end pipe becomes $L_1=\frac{L}{2}$and $L_2=\frac{L}{2}$(after cutting on half) and their fundamental frequencies become $f_1$and $f_2$ respectively.

Step 2. As the pipe is cut into two equal lengths:

We know that, fundamental frequency of closed at one end pipe is,

$$\begin{gathered} f_1=\frac{V}{4L_1} \\ \Rightarrow f_1=\frac{V}{4\left({\displaystyle \frac{L}{2}}\right)} \\ \Rightarrow f_1=\frac{2V}{4L}\dots \left(2\right) \end{gathered}$$

Putting the value of equation $\left(1\right)$ in equation $\left(2\right)$

$$\begin{gathered} \Rightarrow f_1=2f \\ \Rightarrow f_1=2\times 412 \\ \Rightarrow f_1=824Hz \end{gathered}$$

Step 3. The frequency of open at both end pipe is:

We know that, fundamental frequency of both end open pipe is,

$$\begin{gathered} f_2=\frac{V}{2L_2} \\ \Rightarrow f_2=\frac{V}{2\left({\displaystyle \frac{L}{2}}\right)} \\ \Rightarrow f_2=\frac{V}{L} \\ \Rightarrow f_2=\frac{4V}{4L}\dots \left(3\right) \end{gathered}$$

Putting the value of equation $\left(1\right)$ in equation $\left(3\right)$

$$\begin{gathered} \Rightarrow f_2=4f \\ \Rightarrow f_2=4\times 412 \\ \Rightarrow f_2=1648Hz \end{gathered}$$

Hence, the correct option is (B).