A tuning fork resonates with a sonometer wire of length 1 m stretched with a tension of 6 N. When the tension in the wire is changed to 54 N, the same tuning fork produces 12 beats per second with it. The frequency of the tuning fork is ________________ Hz.

Solution

Physics sir · 4 steps

Tension goes up 9 times, so the wire's frequency goes up only 3 times. The beats are that gap.

1. Frequency and tension. For a wire of fixed length, the frequency is f = 1/2L√(T/μ). With L and μ unchanged, f ∝ √T.
2. New wire frequency. At 6 N the wire resonates with the fork, so its frequency equals the fork frequency f. At 54 N the tension is 9 times larger, so the wire's frequency becomes √9 f = 3f.

   3f - f = 12

3. Beats. The beat frequency is the difference between the two frequencies, so 2f = 12.
4. So. The frequency of the tuning fork is f = 6 Hz. That makes the answer 6.

The common slip: 1.5. Taking frequency proportional to T instead of √T makes the new frequency 9f. Then 8f = 12 gives 1.5, not 6.

The idea to keep. On a sonometer, frequency goes as the square root of tension. Scale the frequency first, then subtract the two frequencies to get the beats.

Answer from the NTA final answer key. Solution by PrepWiser.