Two linear SHM’s of equal amplitude and frequencies ω and 2ω are impressed on a particle along the axes x and y respectively. If the initial phase difference between them is π2 the equation of the resultant path followed by the particle is

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x2+Ay2−A22=0

x2+Ay+A2=0

2x2+Ay2−A22=0

x2+Ay−A22=0

answer is A.

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Detailed Solution

⇒x=Asinωt ⇒y=Asin2ωt+π2=Acos2ωt=A1−2sin2ωt ⇒y=A1−2x2A2 ⇒y=A2−2x2A⇒Ay=A2−2x2 ⇒2x2+Ay=A2 ⇒x2+Ay2−A22=0

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