Question

Three waves of equal frequency having amplitude 10 mm, 4 mm and 7 mm arrive at a given point with successive phase difference 90°. The amplitude of the resulting wave in mm is given by

Answer 1

Let the waves be
x1 = 10sin(ωt)
x2 = 4sin(ωt+π/2)
x3 = 7 sin(ωt+π)

Thus,

x2 = 4cos(ωt)
x3 = -7 sin(ωt)
using relation of trigonometric functions in quadrants

Resultant disturbance by superposition theorem is given by
x = x1 + x2 + x3
x = 10sin(ωt) + 4cos(ωt) - 7 sin(ωt)
= 3 sin(ωt) + 4cos(ωt)
Let R be the resultant amplitude
3 = RcosΦ
4 = RsinΦ
x = RcosΦ sin(ωt) + RsinΦ cos(ωt)

=> x = Rsin(ωt + Φ)
so
R = √(3^2 + 4^2)
= 5 μm

otherwise we can think in a way,
take the new two values
i.e
4cos(ωt) =4sin(ωt+π/2) and
3 sin(ωt)
From equation itself , we get an idea that they are in pi/2 phase difference. In such cases we can use Pythagoras theorem by consider the above two amplitudes as the base and height of a rightangled triangle and the hypotenuse is the resultant of these two sides.

so Hypotenuse=Resultant
=R = √(3^2 + 4^2)
= 5 μm

this is a AIIMS 1995 question

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