Question

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 (Speed of sound in steel $= 5\, km\, s^{-1})$

• A. $1.5 \,kHz$
• B. $2 \,kHz$
• C. $2.5 \,kHz$
• D. $3 \,kHz$

Answer 1

Answer: (C)

$2.5 \,kHz$

Answer 2

Here, $L = 100 \,cm = 1 \,m$, $v=5\, km\,s^{-1}=5 \times 10^{3}\,m\, s^{-1}$ As the rod is clamped at the middle, therefore the middle point is a node. In the fundamental mode, the antinode is formed at each end as shown in figure Therefore, the distance two consecutive antinodes = $L=1 \,m$ But the distance between two consecutive antinodes is $\frac{\lambda}{2}$ $\therefore \frac{\lambda}{2} =1\, m$ or $\lambda=2\,m$ The frequency of the fundamental mode is $\upsilon=\frac{v}{\lambda}=\frac{5\times10^{3}\,m\,s^{-1}}{2\,m}$ $=2.5\times10^{3}\, Hz$ $=2.5\, kHz$