A steel rod `2.5 m` long is rigidly clamped at its centre `C` and longitudinal waves are set up on both sides of `C` by rubbing along the rod . Young's modulus for steel ` = 2 xx 10^(11) N//m^(2)` , density of steel ` = 8000 kg//m^(3)`

If the amplitude of the wave at the antinode , when it is vibrating in its fundamental mode is `2 xx 10^(-6) m` , the maximum velocity of a steel particle in its vibration is

Options

A. `1.25 xx 10^(-2) m//s`
B. `1.25 xx 10^(-3) m//s`
C. `1 m//s`
D. `0.12 m//s`

Answer

A

Explanation

In its fundamental mode of vibration , the wavelength is
`lambda_(0) = 4 xx 1.25 m = 5 m`
And the angular frequency
`omega_(0) = 2 pi n_(0) = ( 2 pi xx 5000)/(5) rad//s`
Hence the maximum velocity of the wave at the antinode is
`v_(0) = A omega_(0) = 2 xx 10^(-6) xx ( 2 pi xx 5000)/(5) m//s`
` = 1.25 xx 10^(-2) m//s`