Ncert Exercises & Solutions

For the harmonic travelling wave, $y = 2 \cos 2 π (10 t - 0.008 x + 3.5)$ where x and y are in cm and t is in second. What is the phase difference between the oscillatory motion at two points separated by a distance of:

1. 4m

2. 0.5m

3. $\frac{λ}{2}$

4. $\frac{3 λ}{4}$ (at a given instant of time)

5. What is the phase difference between the oscillation of a particle located at x=100cm, at t = T sec and t=5 sec?

Hint: Use the standard equation of motion.

Given, the wave equation;

y = 2 \cos 2 π (10 t - 0.0080 x + 3.5)

= 2 \cos (20 πt - 0.016 πx + 7 π)

Now, the standard equation of a travelling wave can be written as:

y = acos (ωt - kx + ϕ)

On comparing with the above equation, we get,

a = 2 cm

ω = 20 π rad / s

k = 0.016 π

Step 1: Find the phase difference in each option.

Path difference

= 4 m

Phase difference,

∆ ϕ = \frac{2 π}{λ} \times path difference (∵ \frac{2 π}{λ} = k)

∆ ϕ = 0.016 π \times 4 \times 100

= 6.4 π rad

∆ ϕ = \frac{2 π}{λ} \times (0.5 \times 100) [∵ Path difference = 0.5 m]

= 0.016 π \times 0.5 \times 100

= 0.8 π rad

∆ ϕ = \frac{2 π}{λ} \times (\frac{λ}{2}) = π rad

[∵ Path difference = λ / 2]

∆ ϕ = \frac{2 π}{λ} \times \frac{3 λ}{4} = \frac{3 π}{2} rad

Step 2: Find the phase difference of the particle at t = T sec and t = 5 sec.

T = \frac{2 π}{ω} = \frac{2 π}{20 π} = \frac{1}{10} s

∴ At x = 100 cm,

t = T

ϕ_{1} = 20 πT - 0.016 π (100) + 7 π

= 20 π (\frac{1}{10}) - 1.6 π + 7 π = 2 π - 1.6 π + 7 π . . . . . . (i)

Again, at x=100 cm, t = 5s

ϕ_{2} = 20 π (5) - 0.016 π (100) + 7 π

= 100 π - (0.016 \times 100) π + 7 π

= 100 π - 1.6 π + 7 π . . . . (ii)

∴

From Eqs. (i) and (ii), we get,

∆ ϕ = phase difference = ϕ_{2} - ϕ_{1}

= (100 π - 1.6 π + 7 π) - (2 π - 1.6 π + 7 π)

= 100 π - 2 π = 98 π rad

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