A metallic rod of length 1m is rigidly clamped at its mid point. Longirudinal stationary wave are setup in the rod in such a way that there are two nodes on either side of the midpoint. The amplitude of an antinode is `2 xx 10^(-6) m`. Write the equation of motion of a point 2 cm from the midpoint and those of the constituent waves in the rod, (Young,s modulus of the material of the rod `= 2 xx 10^(11) Nm^(-2)` , density `= 8000 kg-m^(-3)`). Both ends are free.

Explanation

The situation is shown in figure-6.62.

From figure, we can calculate the wavelength as
`lambda = 0.4 m = 40 cm`
Velocityof longitudinal waves in the rod can be given as
`v = sqrt(((Y)/(P)))`
`sqrt(((2 xx 10^(11))/(8000))) = sqrt(((1 xx 10^(8))/(4))) = (10^(4))/(2)= 5000 m/s`
`n = (v)/(lambda) = (5000)/(0.4) = 12500 Hz`
Assuming left end ofthe rod as origin, the equation ofstationary waves is given by
`y = 2Acos (2pix)/(lambda) sin(2pint)`
`y = 2A cos (2pix)/(lambda) sin(2pi xx 12500 t)`
Where amplitude at any instant t is given by
`R = 2 A cos ((2pix)/(0.4))`
x = 0, `R = 2A = 2xx10^(-6)`m
Thus, equation-(6.203) can bewritten as
`y = 2 xx 10^(-6) cos (5pix) sin(25000pit)`
At a point2 cm from mid-point to the right
`x = 50+2 = 52 cm = 0.52m`
`y = 2xx10^(-6)cos(5pi xx 0.52)sin (25000 pit)`
`y = 2 xx 10^(-6) cos(2.6pi) sin (25000pit)`
This is the required equation of stationary waves in the rod. Now we can write the equations of constituents waves in the rod as
`y_(1) = 1 xx 10^(-6) sin(25000 pi t - 5 pi x)`
`y_(2) = 1 xx 10^(-6)sin(25000 pi t + 5 pix)`