A metallic rod of length 1 m is rigidly clamped at its mid-point. Longitudinal stationary waves are set up in the rod in such a way that there are two nodes on either side of the mid-point. The amplitude of an antinode is `2 xx 10^(-6) `m. Write the equation of morion at a point 2 cm from the mid-point and those of constituent waves in the rod. `(Y = 2 xx 10^(11)N//m^(2)` and `rho = 8 xx 10^(3) kg//m^(3))`

Explanation

In rods, like strings, clamped point is a node while the free ends are antinodes. So the situation, in accordance with given conditions is as shown in figure.

Now as distance between two consecutive nodes is `lambda/2` while between a node and an antinode is `lambda/4`.
`4 xx [lambda/2] + 2[lambda/4] = L`, i.e., `lambda = (2 xx 1)/5 = 0.4 m`.......(i)
Also as `Y = 2 xx 10^(11) N//m^(2)` and `rho = 8 xx 10^(3) kg//m^(3)`
`v = sqrt(Y/rho) = sqrt((2 xx 10^(11))/(8 xx 10^(3))) = 5000 m//s`
So, from `v = flambda, f =(v/lambda) =(5000/0.4) = 12500 Hz`.......(ii)
Now if incident and reflected waves along the rod are `y_(1) =A sin(omegat - kx)` and `y_(2) = A sin(omegat + kx + phi)`, resultant wave will be:
`y=y_(1) + y_(2) = A[sin(omegat - kx) + sin(omegat + kx + phi)]`
But as `sinC + sin D = 2sin(C+D)/2 cos (C -D)/2`
`y = 2A cos[kx + phi/2] = "max"=1`, i.e, `phi =0`
and `A_("max") = 2A = 2 xx 10^(-6) m` (given)
So, `y = 2 xx 10^(-6) cos kx sin omegat` Above equation in the light of Eqns. (i) and (ii) reduces to `y = 2 xx 10^(-6) cos 5pi x sin 25000 pi t`..........(iii)
Now as for a point 2 cm from the mid-point`x=(0.60 +- 0.02)`
`y = 2 xx 10^(-6) co s5pi(0.50 +- 0.02) sin 25000 pit`
This is the required result.
Now as `2cosA sinB = sin(A+B)-sin(A-B)` the resultant wave `y = 2 xx 10^(-6) cos (5pix) sin(25000 pit)` can be written as
`y = 10^(-6)[sin(5pix + 25000 pit) - sin(5pix - 25000pit)]`
i.e., `y =10^(-6) sin[25000 pit + 5pix] + 10^(-6) sin[25000 it - 5pix]` [as `sin(-theta) =- sintheta]`
`y_(2) = 10^(-6) sin[25000 pit - 5pix]` representing the required constituent waves.