Question

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 :

• A. 7
• B. 8
• C. 3
• D. 5

Answer 1

Answer: (A)

$l_{1}=0.516\, m, l_{2}=0.491 \, m, T=20\, N $
Mass per unit length, $\mu=0.001 \, kg / m$
Frequency,
$v_{1}=\frac{1}{2 \times 0.516} \sqrt{\frac{20}{0.001}} $
$v_{2} \frac{1}{2 \times 0.491} \sqrt{\frac{20}{0.001}} $
$\therefore $ Number of beats $=v_{1}-v_{2}=7 .$