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JEE Advanced 2006
Exam
JEE Advanced
Year
2006
Subject
Physics
Chapter
Oscillations

JEE Advanced 2006 · Physics · Oscillations

Function x=A ^2 ω t+B ^2 ω t+C ω t ω t represents SHM.

Function x=Asin⁡2ωt+Bcos⁡2ωt+Cx=A \sin ^2 \omega t+B \cos ^2 \omega t+C sin⁡ωtcos⁡ωt\sin \omega t \cos \omega t represents SHM.

Options

  1. AFor any value of A,BA, B and CC (except C=0C=0 )
  2. BIf A=−B,C=2BA=-B, C=2 B, amplitude =∣B2∣=|B \sqrt{2}|
  3. CIf A=B;C=0A=B ; C=0
  4. DIf A=B;C=2BA=B ; C=2 B, amplitude =∣B∣=|B|

Correct answer

B. If A=−B,C=2BA=-B, C=2 B, amplitude =∣B2∣=|B \sqrt{2}|

Step-by-step solution

For A=−BA=-B and C=2BC=2 B X=B \cos 2 ω\omega t+B \sin 2 ω\omega t=2Bsin⁡(2\omegat+π4)ThisisequationofSHMofamplitude2\sqrt{2 B} \sin \left(2 \omegat+\frac{\pi}{4}\right)This is equation of SHM of amplitude \sqrt{2} BIfIfA=BandandC=2 B,then, thenX=B+B \sin 2 ω\omega tThisisalsoequationofSHMaboutthepointThis is also equation of SHM about the pointX=B.Functionoscillatesbetween. Function oscillates betweenX=0andandX=2 Bwithamplitudewith amplitudeB..
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