Simple Harmonic Motion of a Particle

[HZXAHNLfzjSr]

Homework Statement

A particle is oscillating in simple harmonic motion with period 4.5 ms and amplitude 3.0 cm. At time t = 0, the particle is at the equilibrium position. Calculate, for this particle:

a) frequency
b) the angular frequency
c) the maximum speed
d) the maximum acceleration
e) the speed at time t = 1.0 ms

Homework Equations

None

The Attempt at a Solution

a) 220 Hz
b) 1.4 x 10^3 rad/s
c) 42 m/s
d) 5.9 x 10^4 m/s^2I am stuck at (e) and just don't know how to do this.
I was thinking to use the Instantaneous Velocity formula (Vins = omega {square root[(amplitude)^2 - (displacement)^2]} but that isn't giving me the correct answer.

Thanks for your help in advance!

 

[kuruman]

You need to write down an expression that gives you the position x(t) as a function of time. Take the derivative to get the velocity v(t) as a function of time. From that you can easily find the speed at t = 1.0 ms.

 

[HZXAHNLfzjSr]

Thanks, but can you show me the steps?

 

[SiYuan]

x = x0 sin (omega t)

If you differentiate this function with respect to time, you'll get the function of Velocity.

Both equations (Instantaneous and function of time) should give you the same answer, so check your workings again.

 

[rwisz]

For part (e), you can use the formula for velocity in simple harmonic motion: v = ω√(A^2 - x^2), where ω is the angular frequency, A is the amplitude, and x is the displacement from equilibrium position.

At t = 1.0 ms, the displacement from equilibrium position is 3.0 cm (since the particle starts at equilibrium position and oscillates back and forth). So, plugging in the values given in the problem, we get:

v = (1.4 x 10^3 rad/s)√[(3.0 cm)^2 - (3.0 cm)^2]

= 0 cm/s

The speed at t = 1.0 ms is 0 cm/s. This makes sense because at the equilibrium position, the particle momentarily stops before changing direction and moving back towards the opposite direction.

I hope this helps!