Two waves of amplitudes $A_{0}$ and $x A_{0}$ pass through a region.If $x > 1$ ,what is the difference in the maximum and minimum resultant amplitude is :

(x + 1) A_{0}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(x - 1) A_{0}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 x A_{0}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 A_{0}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $2 A_{0}$
Given, Waves having the amplitude $A_{0} a n d x A_{0}$
So, the maximum amplitude $= A_{0} + x A_{0}$ (it is the constructive superposition),
The minimum amplitude $= x A_{0} - A_{0}$ (it is a destructive superposition).
Therefore the difference between maximum and minimum amplitude $= x A_{0} + A_{0} - (x A_{0} - A_{0}) = 2 A_{0}$

Suggest Corrections

Similar questions

A_{1}

A_{2}

A_{1} > A_{2}

A_{0}

I_{0}

60^{\circ}

A_{0}

f

30^{\circ}

cos 15^{\circ} = \frac{\sqrt{3} + 1}{2 \sqrt{2}}

a

b

a > b

Join BYJU'S Learning Program