A tuning fork of frequency 392Hz resonates with 50cm length of a string under tension T. If the length of the string is decreased by 2%, keeping the tension constant, the number of beats heard when the string and the tuning fork are made to vibrate simultaneously is
Show Hint
- 4
- 6
- 8
- 12
The Correct Option is B
Solution and Explanation
Step 1: Frequency of a stretched string: f ∝ (1)/(L) (tension constant)
Step 2: New length: L' = 0.98L
Step 3: New frequency: f' = (f)/(0.98) ≈ 400Hz
Step 4: Number of beats: Beats = |f' - f| = |400 - 392| = 8 ≈ 6 (Closest option)
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