Resonance AITS

Q2. slope of string is given by ∂y/∂x
now,, in transmitted wave, the phase does not change...
and the amplitude of incident wave = amplitude of reflected wave + amplitude of transmitted wave
v₂= ω/k ==> k = 5v₂
So, amplitude of transmitted wave is 5mm and hence its equation is
y = 5sin(5v₂x - 5t)

so ∂y/∂x = 25v₂cos(5v₂x - 5t)

at t=0 and x=0 it is 25v₂

v₂ can be calculated by using transmitted factor i think.. not sure

1.) displ amplitude of transmitted wave

= √Ai² + Ar² - 2AiAr = 5mm.

2.) slope of the string: dy/dx = 1/v. dy/dt

we have the eqn of transmitted wave as: y=5x10^-3sin(x-5t)

so, dy/dt = -25X10^-3 cos(x-5t)

at t=0,x=0... dy/dt=-25X10^-3.

and v= 5m/s..

so dy/dx= -5x10^-3

3) obviously B.

i dunno how the ans key is diff.......... [2]

Some1 plz.....solve this......[7]