What Is the Smallest Value of Time t for This Simple Harmonic Motion?

[PolarClaw]

Hi,

I've been stuck on trying to figure out what I am not seeing in the problem as i just can't seem to get the answer.

A particle moving in simple harmonic motion with a period T = 1.5s passes through the equilibrium point at time t initial = 0 with a velocity of 1.0m/s to the right. At a time t later, the particle is observed to move to the left with a velocity of 0.50m/s. What is the smallest value of the time t?

I just can't see what to do (although i have calculated the angular frequency ( 4(pi)/3).

Thanks in advance.

 

[Andrew Mason]

Hi,

I've been stuck on trying to figure out what I am not seeing in the problem as i just can't seem to get the answer.

A particle moving in simple harmonic motion with a period T = 1.5s passes through the equilibrium point at time t initial = 0 with a velocity of 1.0m/s to the right. At a time t later, the particle is observed to move to the left with a velocity of 0.50m/s. What is the smallest value of the time t?

I just can't see what to do (although i have calculated the angular frequency ( 4(pi)/3).

The position of the particle will be:

$$x = A_0sin(\omega t + \phi)$$

Given the initial condition, x = 0 at t = 0, $\phi = 0$

The speed is given by:

$$\dot x = -A_0\omega cos(\omega t) = v_0cos(\omega t)$$

If $\dot x(t_1) = v_1 = -.5v_0$, then what is $cos(\omega t_1)$?

So from that, you should be able to work out t1.

AM

 

[PolarClaw]

ahh, i had tried that but i differetiated wrong got the answer now thanks.