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QAIPMT-2009-Ph-2177: NEET 2009 Physics - Class 11 AIPMT-2009-Ph-2177 Solution: Each of the two strings of length 51.6 cm and 49.1 cm are…
Each of the two strings of length 51.6 cm and 49.1 cm are tensioned separately by 20 N force. Mass per unit length of both the strings is same and equal to 1 g/m. When both the strings vibrate simultaneously the number of beats is
Multiple Choice Options
- A 7
- B 8
- C 3
- D 5
Step-by-Step Verified Solution
l 1 = 0.516 m, l 2 = 0.491 m, T = 20 N.
Mass per unit length, _ = 0.001 kg/m.
Frequency, _ = 1 2 l T _
_ 1 = 1 2 _ 0.516 20 0.001
_ 2 = 1 2 _ 0.491 20 0.001
_ Number of beats = _ 1 _ _ 2 = 7 .