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Q.

Two waves are simultaneously passing through a string and their equations are: y₁ = A₁ sin k(x – vt), y₂ = A₂sin k(x – vt + x₀). Given amplitudes A₁ = 12 mm and A₂ = 5 mm, x₀ = 3.5 cm and wave number k = 6.28 cm^–1. The amplitude of the resulting wave will be .......... mm.

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a

7 mm

b

6.5 mm

c

9.6 mm

d

8 mm

answer is A.

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Detailed Solution

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Equations are y₁ = A₁sink(x–vt) and y₂ = A₂sin k(x–vt+x₀). we get:

y₁ = 12sin 6.28(x – vt) and

y₂ = 5sin 6.28(x – vt + 3.5)

Phase shift:

$∆ \emptyset = \frac{2 π}{λ} (∆ x)$

= $k (∆ x)$ (where, K= $\frac{2 π}{λ}$ )

= 6.28 x 3.5 = 7π

A_net = $\sqrt{{A_{1}}^{2} + {A_{2}}^{2} + 2 A_{1} A_{1} c o s \emptyset}$

A_net = $\sqrt{12^{2} + 5^{2} + 2 \times 12 \times 5 \cos (7 π)} = \sqrt{144 + 25 - 120} = 7 m m$

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