A metallic rod of length `1m` has one end free and other end rigidly clamped. Longitudinal stationary waves are set up in the rod in such a way that there are total six antinodes present along the rod. The amplitude of an antinode is `4 xx 10^(-6) m`. Young's modulus and density of the rod are `6.4 xx 10^(10) N//m^(2)` and `4 xx 10^(3) Kg//m^(3)` respectively. Consider the free end to be at origin and at `t=0` particles at free end are at positive extreme.
The equation describing displacements of particles about their mean positions is

Options

A. `s-4 xx 10^(6) cos((11 pi)/(2)x)cos(22 pi xx 10^(3)t)`
B. `s=4 xx 10^(-6)cos((11 pi)/(2)x)sin(22 pi xx 10^(3)t)`
C. `s=4 xx 10^(-6)cos(5pi x)cos (20 pi xx 10^(3)t)`
D. `s=4 xx 10^(-6)cos(5 pi x)sin (20 pi xx 10^(3)t)`

Answer

A

Explanation

Speed of wave `v = sqrt((y)/(rho)) = 3 xx 10^(3)` `lamda = (5 lamda)/(2)+(lamda)/(4) rArr lamda = (4l)/(11)`
Frequency `v=(v)/(lamda)=(4 xx10^(3))/((4)/(11)xx1)=11 xx 10^(3) Hz`, Wave Number `K = (2 pi)/(lamda)=(11 pi)/(2)`
(i) Equation of standing wave in the rod `S=A cos kx sin(omega t + phi)` where `A = 4 xx 10^(-6) m`
`because` at `x=0,t=0 rArr S=A rArr A=A rArr cos k(0) sin phi=1 rArr phi=(pi)/(2)`
`S=4xx 10^(-6) cos((11pi)/(2)x)cos(22 pi xx 10^(3)t)`
(ii) Strain `= (ds)/(dx)=-22pi xx 10^(-6) sin ((11 pi)/(2)x)cos(22 pi xx 10^(3)t) because "stress" = Y xx "strain"`
`rArr "stress" = 140.8 xx 10^(4) cos (22 pi xx 10^(3) t)sin ((11pi)/(2) x+pi)`
(iii) Strain at `t = 1s` and `x=(l)/(2)=(1)/(2)m,|(ds)/(dx)|_(x=(l)/(2))^(t=1)=22 pi xx 10^(-6) xx sin ((11 pi)/(4))=11 sqrt(2) pi xx 10^(-6)`.