KVPY Sample Paper KVPY Stream-SX Model Paper-20

Stationary wave over a stretched string is \[y\,(x,t)=0.01\sin 62.8x\cdot \cos 628t.\] If at mid-point of string there is an anti-node, then

A) fundamental frequency = 10 Hz

B) fundamental frequency =15 Hz

C) fundamental frequency = 20 Hz

D) fundamental frequency = 25 Hz

Correct Answer: C

Solution :

From given equation,

\[\omega =2\pi f=628\]

\[\Rightarrow \,\,\,\,\,f\frac{628}{2\pi }=100Hz\]

This may be third, fifth or seventh harmonic as there is an anti-node at mid-point.

So, fundamental frequency may be

\[\frac{100}{3,}\frac{100}{5},\frac{100}{7}......,\]etc.