- Home
- Standard 11
- Physics
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 $= 5\, km\, s^{-1}$)
$1.5$
$2$
$2.5$
$3$
Solution
Here, $\mathrm{L}=100 \mathrm{\,cm}=1 \mathrm{\,m}, \mathrm{v}=5 \mathrm{\,km} \mathrm{s}^{-1}$
$=5 \times 10^{3} \mathrm{\,ms}^{-1}$
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 $=\mathrm{L}=1 \mathrm{\,m}$
But the distance between two consecutive antinodes is $\frac{\lambda}{2}$
$\therefore \frac{\lambda}{2}=1 \mathrm{\,m}$ or $\lambda=2 \mathrm{\,m}$
The frequency of the fundamental mode is
$v=\frac{v}{\lambda}=\frac{5 \times 10^{3} \mathrm{\,ms}^{-1}}{2 \mathrm{\,m}}=2.5 \times 10^{3} \mathrm{\,Hz}=2.5 \mathrm{\,KHz}$
