Each of the two strings of length $51.6 \mathrm{~cm}$ and $49.1 \mathrm{~cm}$ are tensioned separately by…
Each of the two strings of length $51.6 \mathrm{~cm}$ and $49.1 \mathrm{~cm}$ are tensioned separately by $20 \mathrm{~N}$ force. Mass per unit length of both the strings is same and equal to $1 \mathrm{gm}^{-1}$. When both the strings vibrate simultaneously the number of beats is
5
7
8
3
Solution
Key Idea The number of beats will be the difference of frequencies of the two strings.
Frequency of first string $f_1=\frac{1}{2 l_1} \sqrt{\frac{T}{m}}$
$\begin{aligned}
& =\frac{1}{2 \times 51.6 \times 10^{-2}} \sqrt{\frac{20}{10^{-3}}} \\
& =137.03 \mathrm{~Hz}
\end{aligned}$
Similarly, frequency of second string
$\begin{aligned}
& =\frac{1}{2 \times 49.1 \times 10^{-2}} \sqrt{\frac{20}{10^{-3}}} \\
& =144.01
\end{aligned}$
Number of beats $=\mathrm{f}_2-\mathrm{f}_1=144-137$ $=7$ beats