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A particle moving in simple harmonic motion with a period T = 1.5 s passes through the equilibrium point at time t0 = 0 with a velocity of 1.00 m/s to the right. A time t later, the particle is observed to move to the left with a velocity of 0.50 m/s. (Note the change in direction of the velocity.) The smallest possible value of the time t is

User IPherian
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2 Answers

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Final answer:

The smallest possible value of time t for the particle in simple harmonic motion to change the direction of velocity is half the period T, which is 0.75 s.

Step-by-step explanation:

The question is about finding the smallest possible value of time t for a particle in simple harmonic motion (SHM) that changes its velocity direction. The time period T given is 1.5 s, and the particle initially passes through the equilibrium point with a velocity of 1.00 m/s. Later on, it's moving to the left with a velocity of 0.50 m/s. This means the particle has passed through the equilibrium point once again and reached the other extreme before changing its direction.

Since SHM is symmetric about the equilibrium position, the velocity of the particle will be the same at equal distances from the equilibrium position on either side. Given that the particle starts from the equilibrium position (t0 = 0) with maximum velocity, it will take a quarter period (T/4) to reach one extreme and then another quarter period to pass through the equilibrium point again and reach the second extreme. This is where the particle reverses its direction; thus, it completes a half cycle. So we have t = T/2 to go from maximum velocity to the point where it reverses and has half the velocity in the opposite direction.

Hence, the smallest possible value of time t is T/2 which is 1.5 s / 2 = 0.75 s.

User Pisker
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Answer

given,

Time period= T = 1.5 s

If it's moving through equilibrium point at t₀= 0 with v = 1.0 m/s

v_max=1.00 m/s

we know,

v_ max=A ω

v = A sin (ωt)

-0.50= -1.00 sin (ωt)

sin (ωt) = 0.5


\omega t = sin^(-1)(0.5)


(2\pi)/(T)* t =0.524


(2\pi)/(1.5)* t =0.524

t = 0.125 s

we have time period T=1.5 it is the time to complete one oscillation

means from eq to right,then left,then eq,then left,then from right to eq

time taken for left = t/4 = 0.125/4 = 0.375 s

smallest value of time

=0.375 + 0.125

= 0.50 sec

User Stefano Sambruna
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