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Q.

A particle is subjected to two SHMs, X1=A1 sin ωt and X2=A2 sin ωt+π4. The resultant SHM will have an amplitude of:

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a

A1+A22

b

A12+A22

c

A12+A22+2A1A2

d

A12+A222

answer is C.

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Detailed Solution

Resultant amplitude, AA12+A22+2A1A2cos ϕHere        ϕ=π4⇒     A=A12+A22+2A1A2
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