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…
NEET UG 2009 — Physics Waves & Oscillations
2009mcqmedium
Each of the two strings of length 51.6cm and 49.1cm are tensioned separately by 20N force. Mass per unit length of both the strings is same and equal to 1gm−1. When both the strings vibrate simultaneously the number of beats is
Official previous-year question
Held on 30 Apr 2009 · Verified 9 Jul 2026.
Options
A
5
B
7
C
8
D
3
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Solution
Key Idea The number of beats will be the difference of frequencies of the two strings. Frequency of first string f1=2l11mT=2×51.6×10−2110−320=137.03Hz Similarly, frequency of second string =2×49.1×10−2110−320=144.01 Number of beats =f2−f1=144−137=7 beats
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