Question

Each of the two strings of length $51.6 \, cm$ and $49.1 \, cm$ are equally tensioned by $20 \, N$ force. Mass per unit length of both the strings is the same and equal to $1 \, g/m$ . 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)

$\textit{l}_{1} = \text{0.516 m} , l_{2} = \text{0. 491 m,} T = \text{20 N}$ .
Mass per unit length, $\mu = \text{0.001 kg/m}$
$\text{Frequency} \text{,} v = \frac{1}{2 l ⁡} \sqrt{\frac{ T ⁡}{\mu }}$
$ v _{1} = \frac{1}{2 \times 0 \cdot 51 6} \sqrt{\frac{2 0}{0 \cdot 0 0 1}}$
$ v _{2} = \frac{1}{2 \times 0 \cdot 4 9 1} \sqrt{\frac{2 0}{0 \cdot 0 0 1}}$
∴ Number of beats $= \upsilon_{1} - \upsilon_{2} = 7$