Solving Amplitude Problem for Air Track Glider

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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?

Homework Equations

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]

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

 

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

 

[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)

 

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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]

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.

 

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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 into solve for position.

 

[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).

 

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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]

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.

 

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oh that's right! ok. But I am 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]

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

 

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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]

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.

 

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Ok thank you very much for all your help I really appreciate it. I have one more question. Me and my friend have been trying to figure it out but we can't. Could you please help us. Thank you.
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