Oscillation amplitudeWhat is the amplitude of the oscillation?

[kolua]

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?

Homework Equations

E=K+U

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=½mv_f²

Is this the right way to proceed?

 

[Ronie Bayron]

May be this will help you VIBRATION

 

[JBA]

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 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?

Homework Equations

E=K+U

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=½mv_f²

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]

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.