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Question

Physics Question on Oscillations

Function x=Asin2ωt+Bcos2ωt+Csinωtcosωtx = A sin ^2 \omega t + B cos^ 2 \omega t + C sin \omega t cos \omega t represents SHM

A

For any value of A , B and C (except C = 0)

B

If A=B,C=2B,amplitude=B2A= - B,C = 2B, \, \, amplitude \, \, = | B \sqrt 2 |

C

If A=B;C=0A = B ; C = 0

D

If A=b:C=2b,amplitude=BA = b : C = 2b, \, \, amplitude \, \, = |B |

Answer

If A=b:C=2b,amplitude=BA = b : C = 2b, \, \, amplitude \, \, = |B |

Explanation

Solution

For A = - B and C = 2B
X=Bcos2ωt+Bsin2ωt=2Bsin(2ωt+π4)X =B cos 2 \omega t + B sin 2 \omega t = \sqrt 2 B sin\bigg( 2 \omega t +\frac{\pi}{4} \bigg)
This is equation of SHM of amplitude 2B \sqrt 2 B
.
If A = B and C = 2B , then X=B+Bsin2ωtX = B + B sin 2 \omega t
This is also equation of SHM about the point X = B. Function
oscillates between X = Oand X = 2B with amplitude B.