MCQ
Two identical stringed instruments have frequency $100 Hz$. If tension in one of them is increased by $4\%$ and they are sounded together then the number of beats in one second is
- A$1$
- B$8$
- C$4$
- ✓$2$
✓
Answer
Correct option: D.
$2$
d
(d) Frequency of vibration in tight string
$n = \frac{p}{{2l}}\sqrt {\frac{T}{m}} $
(d) Frequency of vibration in tight string
$n = \frac{p}{{2l}}\sqrt {\frac{T}{m}} $
$\Rightarrow n \propto \sqrt T $
==>$\frac{{\Delta n}}{n} = \frac{{\Delta T}}{{2T}} = \frac{1}{2} \times (4\% ) = 2\% $
==> Number of beats = $\Delta n = \frac{2}{{100}} \times n = \frac{2}{{100}} \times 100 = 2$
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