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Simple harmonic motion of a particle

~christina~

Gold Member
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1. Homework Statement

A particle moves along x axis It is initially at 0.270m moving with velocity of 0.140m/s
and acceleration of -0.320m/s^2
Suppose it moves with constant acceleration for 4.50s.

Find it's:
a) position and velocity at end of time interval

next assume it moves with simple harmonic motin for 4.50s and x= 0 is it's equillibrium position.

Find it's:

b) position and velocity at end of this time interval

2. Homework Equations
[tex]x(t)= A cos (\omega t + \phi )[/tex]
[tex]v(t)= - \omega A sin (\omega t + \phi ) [/tex]
[tex]a(t)= - \omega ^2 A cos (\omega t + \phi) [/tex]


3. The Attempt at a Solution

well I know for the first situation that:

[tex]x_o[/tex]= 0.270
[tex]v_o[/tex]= 0.140m/s
[tex]a_o[/tex]= -0.320m/s^2

and the acceration remain constant for 4.50s

to find a) the postiion and velocity at end of time interval....

I was thinking that I would take the period (T) to equal T= 4.50s?
and would I use 4.51 seconds and use the velocity as a function of time equation

[tex]v(t)= - \omega A sin (\omega t + \phi ) [/tex]

though I need the [tex] \phi [/tex]....hm..

and I think I'm lost here.

can someone Please help me out with this?
 
Last edited:
99
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so, a) it's no problem as is an 1 dimension problem, with movement equations...

about b), it's a whole new problem:
you will need to know the position and velocity at [tex]t_0=4.50[/tex]
so, you can write: [tex]t_1=0[/tex] at [tex]x_1=0[/tex](equilibrium point)
by a), you calculated velocity at [tex]t_0[/tex], so, it's the particles velocity at [tex]t_1[/tex]

I guess this should help...
 
Last edited:

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