Question

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

• A. $163.84\mathrm{Hz}$
• B. $400\mathrm{Hz}$
• C. $320\mathrm{Hz}$
• D. $204.8\mathrm{Hz}$

Answer 1

The correct option is B

$400\mathrm{Hz}$

Step 1: Given data

Frequency of first tuning fork = $256\mathrm{Hz}$

Length of first tuning fork = $25\mathrm{cm}$

Length of second tuning fork = $16\mathrm{cm}$

Step 2: Formula used

To find the required frequency of the second fork, we will use the formula given below:

$\mathrm{f}=\frac{1}{2\mathrm{L}}\sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$

$\mathrm{f}\propto \frac{1}{\mathrm{L}}$
Thus, frequency is inversely proportional to the length of the fork.
Step 3: Find the frequency of the second tuning fork

The relation between frequency and length of tuning fork for first tuning fork:

$256=\frac{\mathrm{k}}{25}$ ………………..(1)

For second tuning fork:

$f_2=\frac{\mathrm{k}}{16}$ ……………….(2)

Comparing the two equations, we get:

$\left(\frac{256}{{\mathrm{F}}_2}\right)=\frac{16}{25}$

$f_2=16\times 25=400\mathrm{Hz}$

Hence, option (b) is correct.

The frequency of the second tuning fork is $400\mathrm{Hz}$.