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.
A
The resultant amplitude (1+√2)a
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B
The phase of the resultant motion relative to the first is 90o
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C
The energy associated with the resulting motion is (3+2√2) times the energy associated with any single motion
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D
The resulting motion is not simple harmonic
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Solution
The correct options are A The resultant amplitude (1+√2)a C The energy associated with the resulting motion is (3+2√2) times the energy associated with any single motion Let y1=asin(ωt−π4) y2=asin(ωt) y3=asin(ωt+π4) On super imposing, resulting SHM- y=a[sin(ωt−π4)+sinωt+sin(ωt+π4)] ⟹y=a[2sinωtcosπ4+sinωt] ⟹y=a(1+√2)sinωt ∴ Resultant amplitude =(1+√2)a Also, EresultantEsingle=(Aa)2 ⟹EresultantEsingle=(√2+1)2 ⟹EresultantEsingle=(3+2√2) ∴Eresultant=(3+2√2)Esingle