Waves - Result Question 56
58. 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) 0.02
(b) 0.03
(c) 0.04
(d) 0.01
[2011M]
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Answer:
Correct Answer: 58. (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 $df=6$
$f=600 Hz$
$\therefore \frac{dT}{T}=\frac{2 \times 6}{600}=0.02$
