A steel rod of length 100 , cm is clamped at | vedclass

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(C) Given: Length of the rod $L = 100 \, cm = 1 \, m$. Speed of sound in steel $v = 5 \, km \, s^{-1} = 5 	imes 10^3 \, m \, s^{-1}$.Since the rod is

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$2.5$

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(C) Given: Length of the rod $L = 100 \, cm = 1 \, m$. Speed of sound in steel $v = 5 \, km \, s^{-1} = 5 	imes 10^3 \, m \, s^{-1}$.Since the rod is clamped at the middle,the middle point acts as a node $(N)$. In the fundamental mode of longitudinal vibration,antinodes $(A)$ are formed at both free ends of the rod.As shown in the figure,the total length of the rod $L$ corresponds to the distance between two consecutive antinodes.The distance between two consecutive antinodes is $\frac{\lambda}{2}$.Therefore,$\frac{\lambda}{2} = L = 1 \, m$,which gives the wavelength $\lambda = 2 \, m$.The frequency of the fundamental mode is given by $f = \frac{v}{\lambda}$.Substituting the values: $f = \frac{5 	imes 10^3 \, m \, s^{-1}}{2 \, m} = 2.5 	imes 10^3 \, Hz = 2.5 \, kHz$.

$A$ steel rod of length $100 \, cm$ is clamped at the middle. The frequency of the fundamental mode for the longitudinal vibrations of the rod is ..... $kHz$ (Speed of sound in steel $= 5 \, km \, s^{-1}$)

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