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Questions  

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:

a
A1+A22
b
A12+A22
c
A12+A22+2A1A2
d
A12+A222

detailed solution

Correct option is C

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

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