A copper rod of length `l=50 cm` is clamped at its midpoint. Find the number of natural longitudinal oscillations of the rod in the frequency range from 20 to `50 k Hz`. What are those frequencies equal to ?

Explanation

Since the copper rod is clamped at mid point, it becomes a mode and the two free ends will be antinodes. Thus the fundamental mode formed in the rod is as shown in the Figure .
In this case, `l=(lambda)/( 2)`
So, `v_(0)=(v)/( 2l) =(1)/( 2l) sqrt((E)/(rho))sqrt((E)/( e))`
where `E=` Young's modules and `rho` is the density of the copper .
Similarly the second mode or the first overtone in the rod is as shown in the figure.
Hence `v_(1)=(3v)/( 2l)=(3)/(2l) sqrt((E)/( rho))`
`v=(2n+1)/(2l)sqrt((E)/(rho))`where` n=0,1,2,.....`
Putting the given values of `E` and `rho` in the general equation
`v=3.8(2n+1)kHz`
Hence `v_(0)=3.8 k. Hz, v_(1)=(3.8xx3) k Hz, v_(2)=(3.8)xx5=19kHz,`
`v_(3)=(3.8xx7) =26..6 kHz, v_(4)=(3.8xx9)=34.2kHz,`
`v_(5)=(3.8xx11)=41.8kHx, v_(6)=(3.8)xx13kHz=49kHz` and
`v_(7)=(3.8) xx14kHzgt 50 k Hz. `
Hence the sought number of frequencies between` 20` to ` 50 kHz` equal 4 .