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Question

x=Asin2ωt+Bcos2ωt+Csinωtcosωt represents SHM -

A
For any value of A,B and C (except C=0)
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B
If A=B,C=2B and amplitude =|B2|
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C
If A=B & C=0
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D
If A=B,C=2B and amplitude =|B|
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Solution

The correct option is D If A=B,C=2B and amplitude =|B|
Given,
x=Asin2ωt+Bcos2ωt+Csinωtcosωt

Re writing the above equation we get,

x=(A2+B2)(A2B2)cos2ωt+C2sin2ωt ......(1)

If A=B & C=2B , then from (1) we get that ,

x=Bcos2ωt+Bsin2ωt

By multiplying and dividing with 2 on RHS,

x=2×B×[12cos2ωt+12sin2ωt]

sin(A+B)=sinAcosB+cosAsinB

we can write the above equation as,

x=B2[sin(2ωt+π4)]

This is an expression for particle executing SHM.

Amplitude of SHM=|B2|

If A=B,C=2B, from (1) we can write that,

x=B+Bsin2ωt

Comparing the above equation with x=x0+Asinωt,

We can say that it is also expression for SHM about point x=B.

The function oscillates between x=0 & x=2B with an amplitude B.

If A=B & C=0, we get x=B

This is clearly not an expression for SHM.

If A , B & C can take any values except C=0 then, we can say from (1) that it is an expression of SHM of a particle which can oscillate about any point that depends upon the values of A and B.

Hence, options (a) , (b) and (d) are the correct answers.

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