Q. 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
Explanation
(d) Frequency of vibration in tight string
$n = \frac{p}{2l}\sqrt{\frac{T}{m}} \Rightarrow n \propto \sqrt{T} \Rightarrow \frac{\mathrm{\Delta}n}{n} = \frac{\mathrm{\Delta}T}{2T} = \frac{1}{2} \times (4\%) = 2\%$
$\Rightarrow$ Number of beats $= \mathrm{\Delta}n = \frac{2}{100} \times n = \frac{2}{100} \times 100 = 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.