Determining acceleration of an object in oscillatory motion

  • #1

Homework Statement



A mass is oscillating on a spring with a period of 4.85 s. At t = 0 the mass has zero speed and is at x = 9.90 cm. What is the magnitude of the acceleration at t = 1.40 s?


Homework Equations



a=-w2x
x(t)=Acos(wt+phi)

The Attempt at a Solution


I tried using the equation for acceleration by solving for w in the equation dealing with position, but I don't have the phase change constant, and otherwise I have no idea how to begin going about this problem. I need a push in the right direction!
 
Last edited:

Answers and Replies

  • #2
collinsmark
Homework Helper
Gold Member
2,961
1,432
I tried using the equation for acceleration by solving for w in the equation dealing with position,
The problem statement gives you the period of oscillation, T = 4.85 s.

I'm sure you also know that the angular frequency, ω = 2Πf.

What's the relationship between f and T?
but I don't have the phase change constant, and otherwise I have no idea how to begin going about this problem. I need a push in the right direction!
You have equations for a(t) and x(t) already. Find another equation for v(t). Use your two initial conditions to solve for A and Φ. :wink:
 
  • #3
Ok so I know that w=2pi/T

I also have the equation v=-wAsinwt

But my initial time is 0 so that simply cancels all my work out... Is there some other velocity equation I should be using? Or should I be combining two equations I already have?
 
  • #4
collinsmark
Homework Helper
Gold Member
2,961
1,432
Ok so I know that w=2pi/T
Very nice. :approve:
I also have the equation v=-wAsinwt
almost, but not quite.
v = -ωAsin(ωt + Φ)​
 
  • #5
Okay... but since at t=0 the object has no velocity, doesn't this mean that it is at it's maximum displacement and therefore there is no phase change?
 
  • #6
collinsmark
Homework Helper
Gold Member
2,961
1,432
Okay... but since at t=0 the object has no velocity, doesn't this mean that it is at it's maximum displacement and therefore there is no phase change?
Yes, it certainly does. :approve: Although it could be at the negative of the maximum too, but it doesn't matter too much for this problem.

But not all future problems will be as simple as this one. So if you want to show that the phase offset is zero, you should be able to prove it mathematically.

Start with
v = -ωAsin(ωt + Φ),​
and plug in one of your initial conditions, (t = 0, v = 0)
0 = -ωAsin(0 + Φ),​
Divide both sides of the equation by -ωA.
Take the arcsin of both sides of the equation.

What does that tell you about Φ?

Once you know Φ, you can do a similar operation using the position equation, along with your other initial condition, to mathematically solve for A.
 
  • #7
Okay, in theory this would mean that arcsin of 0 = 0 and plugging that into the position equation I'd end up with 9.9/cos(0)=A

So, my phase change is 0 and my amplitude is 9.9, the initial position of x. However, my prof tells me that 9.9 is not the amplitude of motion... So I am obviously missing something. I don't understand how if it's speed is 0 at t=0 it could possibly not have a phase change.
 
  • #8
collinsmark
Homework Helper
Gold Member
2,961
1,432
Okay, in theory this would mean that arcsin of 0 = 0 and plugging that into the position equation I'd end up with 9.9/cos(0)=A

So, my phase change is 0 and my amplitude is 9.9, the initial position of x. However, my prof tells me that 9.9 is not the amplitude of motion... So I am obviously missing something. I don't understand how if it's speed is 0 at t=0 it could possibly not have a phase change.
Okay, if we really wanted to be explicit, more explicit equations for simple harmonic motion are (where the equilibrium position is not necessarily at x = 0):

x(t) = x0 + Acos(ωt + Φ)
v(t) = -ωAsin(ωt + Φ)
a(t) = 2Acos(ωt + Φ)​

I was assuming that x0 was zero, because if it's not zero, there's not enough information in the problem statement to solve for the final answer.

Is there any piece of information missing in the problem statement? Maybe there's a figure or something explaining how the position x = 9.90 cm fits in relative to the center-point of oscillation? Is there another part to this problem that gives the system's equilibrium position in terms of x?

All I'm saying is that the way the problem statement is presently worded in this thread, there's not enough information to determine what x0 is.
 
  • #9
Actually... that's exactly the way it is written word for word.

However I've just realized the problem... I am corresponding with my prof through e-mail and he thought that we were discussing another problem.... What you guys have said is exactly correct and the answer is 0.04 m/s^2, which is what I derived using the equations you suggested.

Thanks for all the help!
 

Related Threads on Determining acceleration of an object in oscillatory motion

Replies
3
Views
3K
  • Last Post
Replies
1
Views
6K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
7
Views
4K
Replies
11
Views
4K
Top