Question

In an experiment with sonometer, a tuning fork of frequency $256Hz$ resonates with a length of $25cm$ and another tuning fork resonates with a length of $16cm$. The tension of the string remains constant. The frequency of the second tuning fork is

• A. $163.84Hz$
• B. $400Hz$
• C. $320Hz$
• D. $204.8Hz$

Answer 1

The correct option is B

$400Hz$

Step 1: Given data

Frequency of the first tuning fork, $F_1=256Hz$

Length of the first tuning fork, $l_1=25cm$

Length of the second tuning fork, $l_2=16cm$

Step 2: To find

Frequency of the second tuning fork, $F_2=?$

Step 3: Formula used

Frequency of a string,

$F=\frac{1}{2l}\sqrt{\frac{T}{m}}$

where $l$ is the length of the tuning fork, $T$ is the tension in the string, $m$ is the mass of the string.

Step 4: Calculation

We know, the frequency of a string,

$F=\frac{1}{2l}\sqrt{\frac{T}{m}}$

Here,$\sqrt{\frac{T}{m}}$ is constant.

Therefore, $F\propto \frac{1}{l}$

$$\begin{gathered} \frac{F_1}{F_2}=\frac{l_2}{l_1} \\ \Rightarrow \frac{256}{F_2}=\frac{16}{25} \\ \Rightarrow F_2=\frac{25\times 256}{16} \\ \Rightarrow F_2=400Hz \end{gathered}$$

Therefore, the frequency of the second tuning fork is $400Hz$.

Hence, option B is correct.