Waves

JEE Main level, numerical answer, medium

50 of 142

Question 50 of 142

The displacement equations of two interfering waves are given by y1=10sin⁡(ωt+π3)cm{y}_{1}=10\sin \left(\omega t+\dfrac{\pi }{3}\right)cm, y2=5[sin⁡(ωt)+3cos⁡ωt]cm{y}_{2}=5\left[\sin \left(\omega t\right)+\sqrt{3}\cos \omega t\right]cm respectively. The amplitude of the resultant wave is _____ cmcm.

Show answer and solution

Answer: 20

Displacement equation of second wave can be written as,y2=5(sin⁡ωt+3cos⁡ωt)=10(sin⁡ωt×12+32×cos⁡ωt)=10(sin⁡ωt×cos⁡π3+sin⁡π3×cos⁡ωt){y}_{2}=5\left(\sin \omega t+\sqrt{3}\cos \omega t\right) =10\left(\sin \omega t\times \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\times \cos \omega t\right) =10\left(\sin \omega t\times \cos \dfrac{\pi }{3}+\sin \dfrac{\pi }{3}\times \cos \omega t\right)
=10sin⁡(ωt+π3)=10\sin \left(\omega t+\dfrac{\pi }{3}\right)
As we can see, the phase difference between the waves is 00. Therefore, resultant amplitude will be sum of the amplitudes of the waves.
So A=A1+A2=10cm+10cm=20cmA={A}_{1}+{A}_{2}=10cm+10cm=20cm

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Standing waves on a string simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Waves (Interference and Superposition of Waves), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.