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 strings vibrate simultaneously the number of beats is

• A. 7
• B. 8
• C. 3
• D. 5

Answer 1

The correct option is A 7

The number of beats will be the difference of frequencies of the two strings.

frequency of first string

$f_1=\frac{1}{{21}_1}\sqrt{\frac{T}{m}}$

$=\frac{1}{2\times 51.6\times {10}^{-2}}\sqrt{\frac{20}{{10}^{-3}}}$

similarly, frequency of second string

= $\frac{1}{2\times 49.1\times {10}^{-2}}\sqrt{\frac{20}{{10}^{-3}}}$

Number of beats = $f_2-f_1$

= 144 -137

=7 beats