Waves Ques 54
54. 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
[2011 M]
(a) $0.02$
(b) $0.03$
(c) $0.04$
(d) $0.01$
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Answer:
Correct Answer: 54.(a)
Solution:
- (a) For fundamental mode, $f=\frac{1}{2 \ell} \sqrt{\frac{T}{m}}$
Taking logarithm on both sides, we get
$\Rightarrow \log f=\log (\frac{1}{2 \ell})+\log (\sqrt{\frac{T}{\mu}})$
$\Rightarrow \log (\frac{1}{2 \ell})+\frac{1}{2} \log (\frac{T}{\mu})$
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 $df=6$
$f=600$ $ Hz$
$\therefore \frac{dT}{T}=\frac{2 \times 6}{600}=0.02$
BETA
