Two waves, travelling in the same direction through the same region, have equal frequencies, wavelengths and amplitudes. If the amplitude of each wave is 4 mm and the phase difference between the waves is $90^{\circ}$ , what is the resultant amplitude ?

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Solution

Phase difference, $ϕ = \frac{π}{2},$

f and $λ$ are same. So, $ω$ is same

$y_{1} = r s i n w t$

$y_{2} = r s i n (w t + \frac{π}{2})$

From the principle of superposition

$y = y_{1} + y_{2}$

$= r s i n w t + r s i n (w t + \frac{π}{2})$

$= r [s i n w t + s i n (w t + \frac{π}{2})]$

$= r [2 s i n {(\frac{w t + w t + \frac{π}{2}}{2})} c o s {\frac{w t - w t - \frac{π}{2}}{2}}]$

$\Rightarrow y = 2 r s i n (w t + \frac{π}{4}) c o s (- \frac{π}{4}) \Rightarrow y = \sqrt{2} r s i n (w t + (\frac{π}{4}))$

Resultant amplitude

$= \sqrt{2} r = 4 \sqrt{2} m m$

(because r = 4 mm)

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Two waves, travelling in the same direction through the same region, have equal frequencies, wavelengths and amplitudes. If the amplitude of each wave is 4 mm and the phase difference between the waves is 90∘, what is the resultant amplitude ?

Two waves, travelling in the same direction through the same region, have equal frequencies, wavelengths and amplitudes. If the amplitude of each wave is 4 mm and the phase difference between the waves is $90^{\circ}$ , what is the resultant amplitude ?