Question

Two simple harmonic motions are represented by the equation y1 = 0.1 sin (100πt + π/3) and y2 = 0.1 cosπt. The phase difference of the velocity of particle 1 w.r.t. the velocity of the particle 2 is

Answer 1

Given equations of wave are :
y₁ = 0.1sin(100πt + π/3)
y₂ = 0.1cosπt = 0.1sin(π/2 + πt) = 0.1sin(πt + π/2)

Now, for finding velocity of particle , differentiate both equations with respect to time .
dy₁/dt = v₁ = 0.1 × 100πcos(100πt + π/3) = 10πcos(100πt + π/3)
similarly for 2nd equation,
dy₂/dt = v₂ = 0.1 × π cos(πt + π/2) = 0.1πcos(πt + π/2)

We know ,
if equation x = Asin(ωt + Ф) is given then, at t = 0 phase of motion is Ф
Similarly at t = 0 phase of 1st particle velocity is π/3
at t = 0, phase of velocity of 2nd particle is π/2

Now, phase difference = phase of 1st particle at t = 0 - phase of 2nd particle at t = 0
= π/3 - π/2 = -π/6

Hence, answer is -π/6

Answer 2

The solution is provided in the attachment.
PLEASE MARK AS BRAINLIEST