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now i think the image is better...
6 Answers
106Q2. 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
106Q3. as there is no phase difference between incident and reflected wave... (as given in question) hence (c)(d)
11.) 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]