Question:
Two linear SHMs of equal amplitude and frequencies w and 2w are impressed on a particle along x and y-axes respectively. If the initial phase difference between them is 90, find the resultant path followed by the particle.
Answer:
Let the harmonic motions be represented by x = A sin ωt and y = A sin (2ωt + ∏/2 ) = A cos 2ωt
y = A cos 2ωt
= A (1 – 2 sin2 ωt)
= A (1 – 2 (x/A)2 )
y = 1/A(A2 – 2x2) is the resultant path followed by the particle.
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