Ncert Exercises & Solutions

The displacement of an elastic wave is given by the function $y = 3 sinωt + 4 cosωt,$ where y is in cm and t is in seconds. Calculate the resultant amplitude.

Hint: Use the concept of superposition of waves.

Step 1: Find the phase of the resultant wave.

Given, displacement of an elastic wave,

y = 3 sinωt + 4 cosωt,

Assume,

3 = acosϕ . . . . . (i)

4 = asinϕ . . . . . (ii)

On dividing Eq. (ii) by Eq. (i):

\tan (ϕ) = \frac{4}{3} \Rightarrow ϕ = \tan^{- 1} (4 / 3)

Step 2 : Find the amplitude of the resultant wave .

Also, a^{2} \cos^{2} ϕ + a^{2} \sin^{2} ϕ = 3^{2} + 4^{2}

a^{2} (\cos^{2} ϕ + \sin^{2} ϕ) = 25

a^{2} . 1 = 25 \Rightarrow a = 5

Hence, y = 5 cosϕsinωt + 5 sinϕcosωt

= 5 [cosϕsinωt + sinϕcosωt] = 5 \sin (ωt + ϕ)

where ϕ = \tan^{- 1} (4 / 3)

Hence, amplitude = 5 cm

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