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)

7

Answer 2

$l_1=0.516\, m, l_2=0.49 1\, m,\, T=20\, N.$
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 $=v_1-\upsilon_2=7.$