Q. Distance between nodes on a string is 5 cm. velocity of transverse wave is$2ms^{- 1}$. Then the frequency is
A) 5 Hz
B) 10 Hz
C) 20 Hz
D) 15 Hz
Explanation
(c) Distance between modes
$\frac{\lambda}{2} = 5\ cm$
∴ λ=10 cm=0.1n
$\therefore\ \ Frequency\ v = \frac{v}{\lambda} = \frac{2}{0.1} = 20Hz$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.