Three simple harmonic motions in the same direction having the same ampitude a and same period are superposed. If each differs in phase from the next by 45∘, then
A
the resultant amplitude is (1+√2)a
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B
the phase of the resultant motion relative to the first is 90∘
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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 is (1+√2)a C the energy associated with the resulting motion is (3+2√2) times the energy associated with any single motion By principle of superposition
y=y1+y2+y3
=asin(ωt+45∘)+asinωt+asin(ωt−45∘)
=asin(ωt+45∘)+asin(ωt−45∘)+asinωt
=2asinωtcos45∘+asinωt
=√2asinωt+asinωt=(1+√2)asinωt
∴ Amplitude of resultant motion =(1+√2)a
b) The phase of the resultant motion relative to the first is 45∘.
c) Energy is SHM is proportional to square of the amplitude: E∝a2
∴ERE=(1+√2)2a2a2∴ERE=(1+2+2√2)1
or ER=(3+2√2)E
d) Resultant motion is y=(1+√2)asinωt It is a SHM.