# Oscillation amplitudeWhat is the amplitude of the oscillation?

#### kolua

1. Homework Statement
A 1.4-kg cart is attached to a horizontal spring for which the spring constant is 60 N/m . The system is set in motion when the cart is 0.27m from its equilibrium position, and the initial velocity is 2.6 m/s directed away from the equilibrium position.

A. What is the amplitude of the oscillation?
B. What is the speed of the cart at its equilibrium position?

2. Homework Equations
E=K+U

3. The Attempt at a Solution
U=½kΔx2 =½⋅60⋅0.272=2.187J
K=½mv2=½⋅1.4⋅2.62=4.732J
E=K+U=½mvf2

Is this the right way to proceed?

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#### JBA

Gold Member
Your first equations are correct and the K+U for E (the total energy at that point) is correct, but now you need to balance T against the U of the spring at the maximum amplitude of the oscillation.

#### haruspex

Homework Helper
Gold Member
2018 Award
1. Homework Statement
A 1.4-kg cart is attached to a horizontal spring for which the spring constant is 60 N/m . The system is set in motion when the cart is 0.27m from its equilibrium position, and the initial velocity is 2.6 m/s directed away from the equilibrium position.

A. What is the amplitude of the oscillation?
B. What is the speed of the cart at its equilibrium position?

2. Homework Equations
E=K+U

3. The Attempt at a Solution
U=½kΔx2 =½⋅60⋅0.272=2.187J
K=½mv2=½⋅1.4⋅2.62=4.732J
E=K+U=½mvf2

Is this the right way to proceed?
Yes, that will give you the answer to one part. What about the other?

#### kolua

Yes, that will give you the answer to one part. What about the other?
the Velocity at the equilibrium would be, √(2(2.187+4.732)/1.4)=Ve=3.14m/s
then E=½mVe2=½kΔX2, this X here would be the amplitude. X=0.324

#### haruspex

Homework Helper
Gold Member
2018 Award
the Velocity at the equilibrium would be, √(2(2.187+4.732)/1.4)=Ve=3.14m/s
then E=½mVe2=½kΔX2, this X here would be the amplitude. X=0.324
Haven't checked the numbers in detail, but that looks right.