Ncert Exercises & Solutions

15.10 For the travelling harmonic wave $y (x, t) = 2.0 \cos 2 π (10 t - 0.0080 x + 0.35)$

where x and y are in cm and t in sec. Calculate the phase difference between the oscillatory motion of two points separated by a distance of

(a) 4 m,

(b) 0.5 m,

(d) 3λ/4

The equation for a travelling harmonic wave is given as:
$y (x, t) = 2.0 \cos 2 π (10 t - 0.0080 x + 0.35) = 2.0 \cos (20 πt - 0.016 πx + 0.70 π)$
Propagation constant, k = 0.0160π
Amplitude, a = 2 cm
Angular frequency, ω= 20π rad/s
Phase difference is given by:
$ϕ = kx = \frac{2 π}{λ} x$
For x = 4 m = 400 cm,
Φ = 0.016 π × 400 = 6.4 π rad
For x = 0.5 m = 50 cm,
Φ = 0.016 π × 50 = 0.8 π rad
$For x = \frac{λ}{2},$
$ϕ = \frac{2 π}{λ} \times \frac{λ}{2} = π rad$
$For x = \frac{3 λ}{4},$
$ϕ = \frac{2 π}{λ} \times \frac{3 λ}{4} = 1.5 π rad$

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