A rod of length 1.2,m is clamped at the midpo | vedclass

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(D) When a rod of length $L$ is clamped at its midpoint,the midpoint acts as a node,and the two free ends act as antinodes for the fundamental mode of

Vedclass · Accepted Answer

$4.8 	imes 10^6\,m/s$

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(D) When a rod of length $L$ is clamped at its midpoint,the midpoint acts as a node,and the two free ends act as antinodes for the fundamental mode of vibration.For this mode,the length of each half of the rod corresponds to a quarter wavelength,so $L/2 = \lambda/4$,which implies $\lambda = 2L$.The fundamental frequency $f_0$ is given by $f_0 = v/\lambda = v/(2L)$.Rearranging for the wave speed $v$,we get $v = 2L f_0$.Given $L = 1.2\,m$ and $f_0 = 2 	imes 10^6\,Hz$,we substitute these values:$v = 2 	imes 1.2\,m 	imes 2 	imes 10^6\,Hz = 4.8 	imes 10^6\,m/s$.

$A$ rod of length $1.2\,m$ is clamped at the midpoint and its fundamental frequency is $2\,MHz$. What is the speed of the wave inside the rod?

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