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

As the rod is clamped in the middle and three nodes are formed on either side of mid-point.
`therefore` Therefore, we can write `L = (7 lambda)/(2)`
Also length of rod L = 2m
`therefore" "(7 lambda)/(2) = 2`
`lambda = (1)/(7) = 0.14` m
Velocity of longitudinal wave `= upsilon = sqrt((Y)/(rho))`
`= sqrt((2 xx 10^(11))/(6000))`
`= 5.7 xx 10^(3) m//s`
Frequency `= v = (upsilon)/(lambda) = (5.7 xx 10^(3))/(0.14) = 40714.2` Hz
Let A = amplitude of wave.
Amplitude of an antinodes `= 2 xx` amplitude of wave = 2 A
`therefore " "2A = 3 xx 10^(-6) m`
`A = (3)/(2) xx 10^(-6) = 1.5 xx 10^(-6) m`
As the equation of a stationary wave is
x = 4 cm = 0.4 m
As v = 40714.2 Hz
`therefore (1)/(T) = v`
`y = "2 A cos" (2pix)/(lambda) "sin" (2 pi t)/(T)`
Putting values of A, x, `lambda, (1)/(T)`. We get
`y = 2 xx 1.5 xx 10^(-6) "cos" (2pi (0.4))/(0.14) "sin" (2 pi t)/(((1)/(40714.2)))`
`= 3 xx 10^(-6) cos (5.7 pi) sin (81428 pi)`
This is equation of motion of wave at a point 4 cm from the mid-point.
The constituting wave are
`y = A sin ((2pi t)/(T) +- (2pi x)/(lambda))`
`y = 1.5 xx 10^(-6) sin 2pi t(40714 +- (2pi x)/(0.14))`
`= 1.5 xx 10^(-6) sin 2 pi (4.0714 t +- 7.14 x)`