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Two identical piano wires, kept under the same tension T, have a fundamental frequency of 600 Hz. The fractional increase in the tension of one of the wires which will lead to occurrence of 6 beats/s when both the wires oscillate together would be
A
0.04
B
0.01
C
0.02
D
0.03
Detailed Solution
Fundamental frequency of a stretched wire: $f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$, so $f \propto \sqrt{T}$
For a small change: $\frac{\Delta f}{f} = \frac{1}{2}\frac{\Delta T}{T}$
6 beats per second means the frequencies differ by $\Delta f = 6$ Hz.
$\frac{6}{600} = \frac{1}{2}\frac{\Delta T}{T}$
$\frac{\Delta T}{T} = 2\times\frac{6}{600} = \frac{1}{50}$
$\frac{\Delta T}{T} = 0.02$
For a small change: $\frac{\Delta f}{f} = \frac{1}{2}\frac{\Delta T}{T}$
6 beats per second means the frequencies differ by $\Delta f = 6$ Hz.
$\frac{6}{600} = \frac{1}{2}\frac{\Delta T}{T}$
$\frac{\Delta T}{T} = 2\times\frac{6}{600} = \frac{1}{50}$
$\frac{\Delta T}{T} = 0.02$