Thread: Simple harmonic quesion

Hi, I'm not very good at my mechanics, notably simple harmonic motion. I have been given this question:

A particle P moves with simple harmonic motion and comes to rest at two points A and B which are 0.24m apart on a horizontal line. The time for P to travel from A to B is 1.5s.

The midpoint of AB is O. At time t = 0, P is moving through O, towards A, with speed u m/s

a) Find the value of u
b) Find the distance of P from B when t = 2s
c) Find the speed of P with t = 2s

OKAY
So
a) I know T = 2pi/w, as it takes 1.5s to go from A to B, that means T=3

Therefore w = 2pi/3
I know v² = w²(a²-x²)
but I don't know how that helps?

Am I doing it right so far?

EDIT: We are told that at t = 0 x is effectively 0, so:

and v = u

so u =(2pi/3)a

But what's next? How do I find a?
Thanks

Originally Posted by thomas49th

Hi, I'm not very good at my mechanics, notably simple harmonic motion. I have been given this question:

A particle P moves with simple harmonic motion and comes to rest at two points A and B which are 0.24m apart on a horizontal line. The time for P to travel from A to B is 1.5s.

The midpoint of AB is O. At time t = 0, P is moving through O, towards A, with speed u m/s

a) Find the value of u
b) Find the distance of P from B when t = 2s
c) Find the speed of P with t = 2s

OKAY
So
a) I know T = 2pi/w, as it takes 1.5s to go from A to B, that means T=3

Therefore w = 2pi/3
I know v² = w²(a²-x²)
but I don't know how that helps?

(a) "a" is the maximum displacement from equilibrium

v² = (2pi/3)² (.12² - 0²)
v = .25 m/s

note that the maximum speed occurs at the equilibrium position.
max speed can also be found using this equation ... v{max} = aw

(b) let A be at position -.12 m and B at position +.12 m

position function ... x(t) = -.12 sin[(2pi/3)t]

(c) velocity function ... v(t) = -.12(2pi/3) cos[(2pi/3)t]