Question

Each of the two strings of length $51.6cm$ and $49.1cm$ are equally tensioned by $20N$ force. Mass per unit length of both the strings is the same and equal to $1g/m$ . When both the strings vibrate simultaneously in fundamental mode the number of beats is--

• A. 7
• B. 8
• C. 3
• D. 5

Answer 1

Answer: (A)

${\text{l}}_1=\text{0.516 m},l_2=\text{0. 491 m,}T=\text{20 N}$ .
Mass per unit length, $\mu =\text{0.001 kg/m}$
$\text{Frequency}\text{,}v=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$
$v_1=\frac{1}{2\times 0\cdot 516}\sqrt{\frac{20}{0\cdot 001}}$
$v_2=\frac{1}{2\times 0\cdot 491}\sqrt{\frac{20}{0\cdot 001}}$
∴ Number of beats $={\upsilon }_1-{\upsilon }_2=7$