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
- 7
- 8
- 3
- 5
Answer
(1) 7
Solution
Speed of the transverse wave on each string:
v = T = 2010-3 = 2 104 141.4 m/s
Fundamental frequencies = v2L:
1 = 141.42 0.516 137.0 Hz
2 = 141.42 0.491 144.0 Hz
Number of beats per second:
2 - 1 144.0 - 137.0 = 7
Answer: (1)