Question

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
• D. 2

Answer 1

The correct option is D

2

Frequency of vibration in tight string

n = $\frac{P}{2l}\sqrt{\frac{T}{m}}\Rightarrow n\propto \sqrt{T}\Rightarrow \frac{△n}{n}=\frac{△T}{2T}=\frac{1}{2}\times \left(4\%\right)=2\%$

$\Rightarrow Numberofbeats=△n=\frac{2}{100}\times n=\frac{2}{100}\times 100=2$