Q. A string of length 2m is fixed at both ends. If this string vibrates in its fourth normal mode with a frequency of 500 Hz, then the waves would travel on it with a velocity of
A) $125\ ms^{- 1}$
B) $250\ ms^{- 1}$
C) $500\ ms^{- 1}$
D) $1000\ ms^{- 1}$
Explanation
(c) In general, pth mode of a string fixed at ends has frequency
$v = \frac{4v}{2l}$ p=1,2,3….
Where v is velocity of wave and 𝒍 is length of string. In fourth normal mode, p=4
$\therefore\ \ \ \ \ \ v = \frac{4v}{2l}$
Given, v=500 Hz, 𝒍=2m
Hence,
$500 = \frac{4v}{2 \times 2}$
Or
$v = \frac{500 \times 4}{4} = 500ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.