Wave Mechanics Question 649
Question: 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
Options:
A) 0.02
B) 0.03
C) 0.04
D) 0.01
Show Answer
Answer:
Correct Answer: A
Solution:
[a] For fundamental mode, $ f=\frac{1}{2\ell }\sqrt{\frac{T}{\mu }} $ Taking logarithm on both sides, we get $ \log ,f=\log ( \frac{1}{2\ell } )+\log ( \sqrt{\frac{T}{\mu }} ) $
$ =\log ( \frac{1}{2\ell } )+\frac{1}{2}\log ( \frac{T}{\mu } ) $ or $ \log ,f=\log ( \frac{1}{2\ell } )+\frac{1}{2}[\log ,T-\log \mu ] $ Differentiating both sides, we get $ \frac{df}{f}=\frac{1}{2}\frac{dT}{T} $ (as $ \ell $ and $ \mu $ are constants)
$ \Rightarrow \frac{dT}{T}=2\times \frac{df}{f} $ Here $ df=6 $
$ f=600Hz $
$ \
Therefore \frac{dT}{T}=\frac{2\times 6}{600}=0.02 $
BETA
