# Amplitude problem

1. Jun 16, 2009

### Liketothink

1. The problem statement, all variables and given/known data
An air-track glider is attached to a spring. The glider is pulled to the right and released from rest at t=0 s. It then oscillates with a period of 1.5 s and a maximum speed of 40.4 cm/s. What is the amplitude of the oscillation?

2. Relevant equations
x=Acos(wt+ro)

3. The attempt at a solution
I really don't know how to start this problem because angular speed is not given and the length the air glider travels is not given either.

2. Jun 16, 2009

### dx

You can find the angular velocity from the period: T = 2π/ω.

3. Jun 16, 2009

### Liketothink

I found it to be 4.19 rad/s. I don't see how it helps me though.

4. Jun 16, 2009

### dx

The maximum kinetic energy is ½mv2 where v = 40.4 cm/s, at the center where potential energy is zero. Now at the maximum displacement (x = A), all of this KE must be potential energy (KE is zero at this point), so

½mv2 = ½kA2

Write k in terms of ω and m, and solve for A. (m cancels out)

5. Jun 16, 2009

### Liketothink

Ok so v^2/w^2=A^2. I have one question though. Why is it 1/2mv^2 instead of 1/2kx^2 if all the kinetic Energy is transferred to potential energy at the amplitude?

6. Jun 16, 2009

### dx

At x, the potential energy is given by ½kx2. If x = A, then the potential energy is ½kA2. This must be equal to the max kinetic energy, so ½mv2 = ½kA2.

7. Jun 16, 2009

### Liketothink

I see. If I wanted to find the position at t=.5s for that question wouldn't I just take x=Acos(wt+ro)? I have tried that and it seemed to be wrong. I solved for ro when x=0, ro=pi/2 then I solved for x but that is wrong. I tried putting the calculator in degrees or radians mode it's still wrong. What am I doing wrong? What mode is the calculator suppose to be in to solve for position.

8. Jun 16, 2009

### dx

r0 (initial phase) is zero in this case. How did you get π/2?

The position at t = 0.5 would just be x = Acos(0.5ω).

9. Jun 16, 2009

### Liketothink

I set x=0 so that 0=Acos(wt+ro). Since t=0, 0=Acos(ro). Then 0=cos(ro), so I get ro=pi/2.

10. Jun 16, 2009

### dx

The glider is pulled to the right and released from rest at t = 0, so at t = 0 the positin is x = A.

A = Acos(r0)

cos(r0) = 1

so r0 = 0.

11. Jun 16, 2009

### Liketothink

oh that's right! ok. But Im still getting the wrong answer though. x=.096cos(4.19*.5)=
.0963 m. The computer is telling me it's wrong. what am I doing wrong in this case?

12. Jun 16, 2009

### dx

Hm.. maybe it wants you to enter the answer in cm?

13. Jun 16, 2009

### Liketothink

I got it. For some reason the answer was -.042 when the calculator was in radian. do you know why I have to put the calculator in radians? I have a test tomorrow and I would not have gotten this answer right.

14. Jun 16, 2009

### dx

Well you calculated the angular velocity in radians, so you must have the calculator in radian mode. If you want to use degrees, just multiply the angular velocity by (360)/2π.

15. Jun 16, 2009