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MHT CET 202625 AprilEvening Shift
Exam
MHT CET
Year
2026
Date
25 April
Shift
Evening Shift
Subject
Physics
Chapter
Oscillations

MHT CET 2026 · 25 April · Evening Shift · Physics · Oscillations

Two SHM's x_1 = a (ω t + (3π)/(2)) and x_2 = a ( ω t + π) are...

Two SHM's x1=asin⁡x_1 = a\sin\left(\omegat+t +\dfrac{3\pi}{2}\right)andand x_2 = a\sin(ω\omega t + \pi) are superimposed on each other. The resultant amplitude of motion is (sin⁡(\sin\pi/2=1/2 = 1 and \cos \pi/2=0)/2 = 0)

Options

  1. A2a2a
  2. B2a22a^2
  3. Ca2a\sqrt{2}
  4. D00

Correct answer

C. a2a\sqrt{2}

Step-by-step solution

Given equations of SHM are x1=asin⁡x_1 = a\sin\left(\omegat+t +\dfrac{3\pi}{2}\right)andx2=asin⁡(and x_2 = a\sin(\omegat+π).<br><br>ThephasedifferencebetweenthetwoSHMsisΔϕ=3π2−π=π2.<br><br>Theresultantamplitudet + \pi).<br><br>The phase difference between the two SHMs is \Delta\phi = \dfrac{3\pi}{2} - \pi = \dfrac{\pi}{2}.<br><br>The resultant amplitude AisgivenbyA=a12+a22+2a1a2cos⁡(Δϕ).<br><br>Substitutingis given by A = \sqrt{a_1^2 + a_2^2 + 2a_1a_2\cos(\Delta\phi)}.<br><br>Substituting a_1 = a,,a_2 = a,andΔϕ=π2:<br><br>, and \Delta\phi = \dfrac{\pi}{2}:<br><br> A = a2+a2+2a2cos⁡(π2)<br><br>Sincecos⁡(π2)=0\sqrt{a^2 + a^2 + 2a^2\cos\left(\dfrac{\pi}{2}\right)}<br><br>Since \cos\left(\dfrac{\pi}{2}\right) =0, we get:

A = \sqrt{2a^2} = a\sqrt{2}<br><br>Answer:<br><br>Answer:a\sqrt{2}
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