Spring Constant and Oscillations -

[mparsons06]

Homework Statement

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.

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

 

[zachzach]

Homework Statement

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.

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

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 <br /> <br /> \omega = {\sqrt{\frac{k}{m}} \rightarrow x(t) = Asin({\sqrt{\frac{k}{m}}t)

 

[mparsons06]

Can you help me with part b?

 

[zachzach]

Can you help me with part b?

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

 

[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?

 

[zachzach]

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

 

[zachzach]

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?

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.

 

[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

 

[zachzach]

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

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. :/

 

[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.

 

[mparsons06]

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

 

[zachzach]

That's what I get.