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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 by

y1=asinωtπ4;y2=asinωt and y3=asinωt+π4

 Resultant will be y=asinωtπ4+sinωt+sinωt+π4

=a2sinωtcosπ4+sinωt=a[2sinωt+sinωt]=a(1+2)sinωt

Resultant amplitude =(1+2)a

Energy is S.H.M. ( Amplitude )2

EResultant ESingle =Aa2=(2+1)2=(3+22)EResultant =(3+22)ESingle 

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