Q. 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 string vibrate simultaneously the number of beats is
A) 5
B) 7
C) 8
D) 3
Explanation
(b) Using $n = \frac{1}{2l}\sqrt{\frac{T}{m}}$
Number of beats $= \frac{1}{2}\sqrt{\frac{T}{m}}\left\lbrack \frac{1}{l_{2}} - \frac{1}{l_{1}} \right\rbrack$
$= \frac{1}{2}\sqrt{\frac{20}{1 \times 10^{- 3}}}\left\lbrack \frac{1}{49.1 \times 10^{- 2}} - \frac{1}{51.6 \times 10^{- 2}} \right\rbrack = 7$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.