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A particle is subjected to two SHM’s $x_{1} = A_{1} S i n ω t a n d x_{2} = A_{2} S i n (ω t + π / 4)$ . The resultant SHM will have an amplitude of

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(A1 + A2)/2

A12+A22

A12+A22+2A1A2

A1A2

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detailed solution

Correct option is C

∅=π4 The resultant SHM will have an amplitude=A=A12+A22+2A1A2cos∅ given phase difference ∅=π4 =A12+A22+2A1A2cosπ4 =A12+A22+2A1A212 resultant amplitude=A12+A22+2A1A2

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A particle is subjected to two SHM’s x1 = A1 Sinωt and x2 = A2 Sin(ωt + π/4). The resultant SHM will have an amplitude of

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A particle is subjected to two SHM’s $x_{1} = A_{1} S i n ω t a n d x_{2} = A_{2} S i n (ω t + π / 4)$ . The resultant SHM will have an amplitude of