SHMHard
Question
Three simple harmonic motion in the same direction having the same amplitude and same period are suspended. If each differ in phase from the next by 45o, then
Options
A.the resultant amplitude is ( 1 + √2)α
B.the phase of the resultant motion relative to the first is 90o
C.the energy associated with the resulting motion is (3 + 2 √2) times the energy associated with any single motion
D.(d) the resulting motion is not simple harmonic
Solution
From superposition priciple
y = y1 + y2 + y3
= a sin ωt + a sin (ωt + 45o) + a sin (ωt + 90o)
= a[ sin ωt + sin (ωt + 90o)] + a sin (ω + 45o)
= 2a sin (ωt + 45o) cos 45o + a sin (ωt + 45o)
= (√2 + 1) a sin (ωt + 45o) = A sin (ωt + 45o)
Therefore, resultant motion is simple harmonic of amplitude
A = (√2 + 1)a
and which differ in phase by 45o relative to the first.
Energy in SHM ∝ (amplitude)2 [E =
mA2 ω2]
∴
= 
= (√2 + 1)2 = (3 + 2√2)
∴ Eresultant = (3 + 2√2) Esingle
y = y1 + y2 + y3
= a sin ωt + a sin (ωt + 45o) + a sin (ωt + 90o)
= a[ sin ωt + sin (ωt + 90o)] + a sin (ω + 45o)
= 2a sin (ωt + 45o) cos 45o + a sin (ωt + 45o)
= (√2 + 1) a sin (ωt + 45o) = A sin (ωt + 45o)
Therefore, resultant motion is simple harmonic of amplitude
A = (√2 + 1)a
and which differ in phase by 45o relative to the first.
Energy in SHM ∝ (amplitude)2 [E =
mA2 ω2]∴
= = (√2 + 1)2 = (3 + 2√2)∴ Eresultant = (3 + 2√2) Esingle
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