Question

A tuning fork produces \(6\) beats per second with both, \(60\ \text{cm}\) and \(40\ \text{cm}\) of the stretched wire of sonometer. The frequency of fork is

• A. \(60\ \text{Hz}\)
• B. \(40\ \text{Hz}\)
• C. \(30\ \text{Hz}\)
• D. \(10\ \text{Hz}\)

Hint:

Frequency is inversely proportional to the wire length. The fork frequency lies between the two wire frequencies, 6 Hz above the lower and 6 Hz below the higher.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
For a sonometer wire under fixed tension, \(f\propto\dfrac{1}{l}\). The shorter wire (40 cm) has the higher frequency.

Step 2: Set up
Let the 60 cm wire have frequency \(f_{1}\). Then the 40 cm wire has \(f_{2}=\dfrac{60}{40}f_{1}=1.5f_{1}\).

Step 3: Use equal beats
The fork gives 6 beats with both, so it lies between them: \(f_{fork}=f_{1}+6=f_{2}-6\).

Step 4: Calculation
\(f_{1}+6=1.5f_{1}-6\), so \(0.5f_{1}=12\) and \(f_{1}=24\ \text{Hz}\). Then \(f_{fork}=24+6=30\ \text{Hz}\).

Final Answer:
The frequency of the fork is 30 Hz. \[ \boxed{\text{(C) }\text{30 Hz}} \]