Question

A sonometer wire of length $25\text{ cm}$ vibrates in unison with a tuning fork. When its length is decreased by $1\text{ cm}$, $6$ beats are heard per second. What is the frequency of the tuning fork?

• A. $200\text{ Hz}$
• B. $72\text{ Hz}$
• C. $100\text{ Hz}$
• D. $144\text{ Hz}$

Hint:

You can set this up instantly using ratios: $\frac{n_2}{n_1} = \frac{l_1}{l_2} \implies \frac{n+6}{n} = \frac{25}{24}$. The fractional gap on the length side is $\frac{1}{24}$. Since the frequency gap must scale exactly to match ($6\text{ Hz}$), the base frequency is simply $24 \times 6 = 144\text{ Hz}$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
Initially, a sonometer string of length $l_1 = 25\text{ cm}$ vibrates at the exact same fundamental frequency as a tuning fork ($n$). This state is called unison.
When the wire's length is shortened to $l_2 = 25 - 1 = 24\text{ cm}$, its fundamental frequency shifts higher, generating a beat frequency of $6\text{ beats/s}$ with the fork. We need to calculate the fork's structural frequency $n$.

Step 2: Key Formula or Approach:
The fundamental frequency of a stretched sonometer wire under stable tension is inversely proportional to its length:
$$n = \frac{v}{2l} \implies n \propto \frac{1}{l} \implies n_1 l_1 = n_2 l_2$$ The beat frequency produced is the positive difference between the two acoustic frequencies: $f_{beat} = |n_2 - n_1|$.

Step 3: Detailed Explanation:
Initially, the wire matches the fork: $n_1 = n$.
Shortening a string limits its wavelength, forcing its structural frequency to rise. Therefore, the modified wire frequency $n_2$ is higher than the fork's frequency:
$$n_2 = n + 6$$ Applying the inverse-proportional relationship formula ($n_1 l_1 = n_2 l_2$):
$$n \times 25 = (n + 6) \times 24$$ Expand the algebraic distribution on the right side:
$$25n = 24n + 144$$ Isolate the unknown value $n$ by shifting $24n$ to the left side:
$$25n - 24n = 144 \implies n = 144\text{ Hz}$$

Step 4: Final Answer:
The frequency of the tuning fork is $144\text{ Hz}$, which matches option (D).