A travelling harmonic wave is given as: y (x, t) = 2.0 cos 2π (10t – 0.0080x + 0.35).

What is the  phase difference between the oscillatory motion of two points separated by a distance of:

(i) 8 m,

(ii) 1 m,

(iii) $$\dfrac{λ}{2}$$,

(iv) $$\dfrac{6λ}{4}$$

[ X and y are in cm and t is in secs ].

Asked by Pragya Singh | 1 year ago |  213

##### Solution :-

Given,
Equation for a travelling harmonic wave :
y (x, t) = 2.0 cos 2π (10t – 0.0080x + 0.35)
= 2.0 cos (20πt – 0.016πx + 0.70 πWhere,
Propagation constant, k = 0.0160 π
Amplitude, a = 2 cm
Angular frequency, ω= 20 π rad/s

We know,
Phase difference Φ = kx = $$\dfrac{2π}{λ}$$

(i) For x = 8 m = 800 cm
Φ = 0.016 π × 800 = 12.8 π rad

(ii) For x =  1 m = 100 cm
Φ = 0.016 π × 100 = 1.6 π rad

(iii) For x =$$\dfrac{λ}{2}$$
Φ = ($$\dfrac{2π}{λ}$$)  x  ( $$\dfrac{λ}{2}$$)

(iv) For x =$$\dfrac{6λ}{4}$$
Φ = ($$\dfrac{2π}{λ}$$)  x  ( $$\dfrac{6λ}{4}$$)

Answered by Abhisek | 1 year ago

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