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Three simple harmonic motions in the same direction having the same amplitude a and same period are superposed. If each differs in phase from the next by 45o, then

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a
The resultant amplitude is 2a
b
The phase of the resultant motion relative to the first is 90o
c
The energy associated with the resulting motion is (3+22) times the energy associated with any single motion
d
The resulting motion is not simple harmonic

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detailed solution

Correct option is C

Let simple harmonic motions be represented byy1=asin⁡ωt−π4;y2=asin⁡ωt and y3=asin⁡ωt+π4 Resultant will be y=asin⁡ωt−π4+sin⁡ωt+sin⁡ωt+π4=a2sin⁡ωtcos⁡π4+sin⁡ωt=a[2sin⁡ωt+sin⁡ωt]=a(1+2)sin⁡ωtResultant amplitude =(1+2)aEnergy is S.H.M. ∝( Amplitude )2∴EResultant ESingle =Aa2=(2+1)2=(3+22)⇒EResultant =(3+22)ESingle


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