1. Apr 28, 2010

### mparsons06

1. The problem statement, all variables and given/known data

A 1.450 kg air-track glider is attached to each end of the track by two coil springs. It takes a horizontal force of 0.700 N to displace the glider to a new equilibrium position, x= 0.290 m.

a.) Find the effective spring constant of the system.

b.) The glider is now released from rest at x= 0.290 m. Find the maximum x-acceleration of the glider.

c.) Find the x-coordinate of the glider at time t= 0.490T, where T is the period of the oscillation.

d.) Find the kinetic energy of the glider at x=0.00 m.

3. The attempt at a solution

a.) 2.41 N/m

b.) v = A $$\sqrt{k/m}$$
v = 0.290 $$\sqrt{2.41/1.450}$$
v = 0.374 m/s^2

Incorrect. Where did I mess up?

c.) x(t) = 0.290 * sin (w*t)
t = 0.490T
T = 2t * pi / w = 0.98 * pi / w

x(t) = 0.290 sin (w * 0.98 pi / w)
x(t) = 0.290 sin (0.98 pi)
x(t) = - 0.0156 m

Incorrect. Help?

d.) 0.102 J

2. Apr 28, 2010

### zachzach

I do not see why you need to use T here. $$T = \frac{2t\pi}{\omega}$$
is wrong since $$T = \frac{2\pi}{\omega}$$

I would just use $$\omega = {\sqrt{\frac{k}{m}} \rightarrow x(t) = Asin({\sqrt{\frac{k}{m}}t)$$

3. Apr 28, 2010

### mparsons06

Can you help me with part b?

4. Apr 28, 2010

### zachzach

No muiltiplied, I just plugged it into the general form
$$x(t) = Asin(\omega t)$$

5. Apr 28, 2010

### mparsons06

So I did:

x(t) = A sin (sqrt (k/m) * t)
x(t) = (0.290) sin(sqrt (2.41 / 1.450) * 0.490T)
x(t) = (0.290) sin(1.29 * 0.490T)
x(t) = (0.290) sin(0.6321T)
x(t) = (0.290)(0.011)
x(t) = -0.00319 m

Does that seem correct to you?

6. Apr 28, 2010

### zachzach

You know the equation for x(t). Find a(t) and figure out its maximum value.

7. Apr 28, 2010

### zachzach

$$clickme$$" see the tex in brackets? You must put that before you write in latex in brackets and after the last line put /tex in brackets (every time you write in latex). No it is not your calculator is in degrees mode and it should be in radians. Plus your answer is negative for an unknown reason.

Also, you asked if it was divided by t earlier. You know that when you take the sine of something it has to be dimensionless so you could figure out if that is even possible with a little dimensional analysis.

8. Apr 28, 2010

### mparsons06

I changed my calculator to radians:

x(t) = A sin (sqrt (k/m) * t)
x(t) = (0.290) sin(sqrt (2.41 / 1.450) * 0.490T)
x(t) = 0.145 m

9. Apr 28, 2010

### zachzach

I put in those same numbers and got a different answer. And what is T? O wow I misread the problem I didn't notice what that t = 0.49T. :/

10. Apr 28, 2010

### zachzach

You substituted into sin for w when you should have substituted for t (in the original post).

$$\omega = \frac{2\pi}{T} \rightarrow x(t) = Asin(\frac{2\pi}{T}t)$$

Just plug in t and since it is in terms of T, the T's will cancel.

11. Apr 28, 2010

### mparsons06

So I got 0.0182 with my calculator in radians... Is that correct?

12. Apr 28, 2010

### zachzach

That's what I get.