Question

the displacement of two particles executing shm on the same line are given as y1=asin(π/2t+Φ) and y2=bsin(2π/3t+Φ) find phase difference between them at t=1​

Answer 1

In 1st case if we put t=1 it will give the theta as $\pi$/2 + Phi

In second case if we put t=1 it will give the theta as 2$\pi$/3+phi.

Subtracting it both we will get phi cancel and 2$\pi$/3-$\pi$/2.

It will be $\pi$/6 that is 30 degrees.

Answer 2

phase difference between them is π/6

when a particle executing simple harmonic motion, equation of motion of particle is represented by, y = Asin(ωt + Φ)

where A is amplitude, ω is angular frequency and (ωt + Φ) is phase of equation of motion of particle at time t.

here, two particles executing shm on the same line are given as y1 = asin(π/2t + Φ)

y2 = bsin(2π/3t + Φ)

at t = 1s,

phase of 1st particle = (π/2 + Φ)

phase of 2nd particle = (2π/3 + Φ)

now, phase difference between them = (2π/3 + Φ) - (π/2 + Φ)

= 2π/3 - π/3 = π/6

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