Deepak Patra
Last Activity: 10 Years ago
Sol. f = 100 Hz, λ = 2 cm = 2 * 10 base 2 m
∴ wave speed, v = f λ = 2 m/s
a) in 0.015 sec 1^ST wave has travelled
x = 0.015 * 2 = 0.03 m = path diffn
∴ corresponding phase difference, ∅ = 2π*/λ = {2π/ (2 * 10^2)} * 0.03 = 3π.
b) Path different x = 4 cm = 0.04 m
⇒ ∅ = (π/λ)x = {(2π/2 * 10^2) * 0.04} = 4 π.
C) The waves have same frequency, same wavelength and same amplitude
Let, Y base 1 = r sin wt y base 2 = r sin (wt + ∅)
⇒ Y = Y base 1 + y base 2 = [r sin wt + (wt + ∅)
= 2r sin (wt + ∅/2) cos (∅/2)
∴ resultant amplitude = 2r cos ∅/2
So when ∅ = 3π, r = 2 * 10^ 3 m
R base res = 2 * (2 * 10^3) cos (3π/2) = 0
Again, when ∅/2π, R base res = 2 * (2 * 10^3) cos (4π/2) = 4mm.