Question

For the travelling harmonic wave
y (x, t) = 2.0 cos 2π (10t – 0.0080x + 0.35)
Where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of
(a) 4 m,
(b) 0.5 m,
(c) λ/2
(d) 3λ/4
​

Answer 1

Explanation:

Equation for a travelling harmonic wave is given as:

y (x, t) = 2.0 cos 2π (10t – 0.0080x + 0.35)

= 2.0 cos (20πt – 0.016πx + 0.70 π)

Where,

Propagation constant, k = 0.0160 π

Amplitude, a = 2 cm

Angular frequency, ω= 20 π rad/s

Phase difference is given by the relation:

Φ = kx = 2π/λ

(a) For x = 4 m = 400 cm

Φ = 0.016 π × 400 = 6.4 π rad

(b) For 0.5 m = 50 cm

Φ = 0.016 π × 50 = 0.8 π rad

(c) For x = λ/2

Φ = 2π/λ × λ/2 = π rad

(d) For x = 3λ/4

Φ = 2π/λ × 3λ/4 = 1.5π rad.

Answer 2

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Y = 2cos2π(10t - 0.008x + 0.35)

= 2cos(20πt - 0.016πx + 0.7π)

Compare with

Y = acos(wt - Kx + ∅)

w = 20π

K = 0.016π

a = 2 cm

∅ = 0.7π

We know,

∆∅ = 2π/lemda × ∆x

= K.∆x

(a) for ∆x = 4 m = 400 cm

Phase difference (∆∅) = K × 400

∆∅ = 0.016π × 400

= 6.4 rad

(b) for ∆x = 0.5 m = 50 cm

Similarly ,

∆∅ = 0.016π × 50

= 0.8π rad

(c) ∆x = lemda/2

∆∅ = 2π/lemda × lemda/2

= π rad

(d) for ∆x = 3lemda/4

∆∅ = 2π/lemda × 3lemda/4

= 3π/2 rad