# Amplitude problem

## Homework Statement

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?

x=Acos(wt+ro)

## 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.

dx
Homework Helper
Gold Member
You can find the angular velocity from the period: T = 2π/ω.

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

dx
Homework Helper
Gold Member
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)

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?

dx
Homework Helper
Gold Member
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.

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.

dx
Homework Helper
Gold Member
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).

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.

dx
Homework Helper
Gold Member
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.

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?

dx
Homework Helper
Gold Member
Hm.. maybe it wants you to enter the answer in cm?

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.

dx
Homework Helper
Gold Member
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.