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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
3
B
5
C
7
D
8
Detailed Solution
Fundamental frequency of a stretched string: $f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$
T = 20 N, $\mu$ = 1 g/m = $10^{-3}$ kg/m: $\sqrt{\frac{T}{\mu}} = \sqrt{\frac{20}{10^{-3}}} = \sqrt{2\times10^4} = 141.4$ m/s
$f_1 = \frac{141.4}{2\times0.516} = 137.0$ Hz
$f_2 = \frac{141.4}{2\times0.491} = 144.0$ Hz
Number of beats per second = $f_2 - f_1 = 144 - 137$
= 7
T = 20 N, $\mu$ = 1 g/m = $10^{-3}$ kg/m: $\sqrt{\frac{T}{\mu}} = \sqrt{\frac{20}{10^{-3}}} = \sqrt{2\times10^4} = 141.4$ m/s
$f_1 = \frac{141.4}{2\times0.516} = 137.0$ Hz
$f_2 = \frac{141.4}{2\times0.491} = 144.0$ Hz
Number of beats per second = $f_2 - f_1 = 144 - 137$
= 7