# Finding Amplitude

1. Jun 16, 2009

### Liketothink

1. The problem statement, all variables and given/known data
An air-track glider attached to a spring oscillates between the 9.23 cm mark and the 50.23 cm mark on the track. The glider completes 27 oscillations in 44.7 s. angular frequency of the glider? (in rad/s)
3.7933
what is the amplitude of the glider?

2. Relevant equations

x=Acos(wt+ro)

3. The attempt at a solution
I considered that at x=9.23 the amplitude was zero and so is ro.
Hence, x=50.23 would equal just 50.23 cm=Acos(3.79(44.7)). But A turned out to be negative. What am I doing wrong? How do I solve this problem? Thank you in advance.

2. Jun 16, 2009

### dx

What is the center of oscillation here? The amplitude is the maximum distance from the center that the glider reaches.

3. Jun 16, 2009

### Liketothink

Would it be half the distance the glider travels which is 20.5 cm?

4. Jun 16, 2009

### dx

Correct.

5. Jun 16, 2009

### Liketothink

It still gives me a negative number if I do .205/((cos(3.79*44.7)).

6. Jun 16, 2009

### dx

Negative number for what?

7. Jun 16, 2009

### Liketothink

For the amplitude A.

8. Jun 16, 2009

### dx

You just said the amplitude was 20.5 cm, which is correct.

9. Jun 16, 2009

### Liketothink

OH! Ok I get it. Thank you very much.

10. Jun 16, 2009

### Liketothink

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