Amplitude and Vibrations in Springs

[RoyceB]

Homework Statement

The amplitude of vibration of a mass on a horizontal spring experiencing SHM is = 0.13m. The mass is 85g and the force constant is 55N/m. a) What is the maximum elastic potential energy of the system. b) Find the speed of the mass when it's position is x = 7.4cm from the equilibrium point. c) What is it's maximum speed?

A = 0.13m
m = 0.085Kg
k = 55N/m
Δx = 0.074m

Homework Equations

Ee = ½k(Δx)²
Ek = ½mv²
Fx = -kΔx
T = 2π√(m/k)
Fc = mv² / A[/B]

The Attempt at a Solution

a)

For this I simply used the Ee equation and solved for it. I used 0.13m as the amplitude as I figured this is the furthest the spring could stretch out.
Ee = ½k(Δx)²[/B]
Ee = ½55N/m(0.13)²
Ee = 0.46J

b)

This one I solved for T, used the force equation and the Fc equation to solve for speed.

T = 2π√(m/k)
T = 2π√(0.086/55)
T = 0.25s

Fx = -kΔx
Fx = -55(0.074)
Fx = 4.07N

Fc = mv² / A
Fc = (0.086)v² / 0.13
v = 2.5m/s

c) I am kind of lost and thought of making Ee = Ek but would like some assistance.

Also if you could proof check my answers and tell me any errors as I assume my entire b) part maybe wrong.

 

[haruspex]

Your answer for b is a bit off. Probably rounding errors. You could retain more digits during the calculation, but I would solve it using energy. How much elastic PE is lost in going from max extension to 7.4cm.?

Fc = (0.086)v2 / 0.13
v = 2.5m/s

c) I am kind of lost and thought of making Ee = Ek but would like some assistance.

If you mean that the max of the one equals the max of the other, yes. Can you turn that into an equation for the max speed?

 

[RoyceB]

Your answer for b is a bit off. Probably rounding errors. You could retain more digits during the calculation, but I would solve it using energy. How much elastic PE is lost in going from max extension to 7.4cm.?

If you mean that the max of the one equals the max of the other, yes. Can you turn that into an equation for the max speed?

Well for b what if I made Ee = Ek' + Ee' and if Ee on the left side is equilibrium then the it could equal and I could move the other one over and solve for that point. Basically making Ek = Ee.

For c if it is at it's furthest point and highest speed could it be Ee = Ek but at a greater distance for x, like the 0.13m?

 

[haruspex]

Well for b what if I made Ee = Ek' + Ee' and if Ee on the left side is equilibrium then the it could equal and I could move the other one over and solve for that point. Basically making Ek = Ee.

For c if it is at it's furthest point and highest speed could it be Ee = Ek but at a greater distance for x, like the 0.13m?

Since you have not defined exactly what you mean by those different energies, I cannot be sure whether you have that right.
It cannot be both at its furthest point and highest speed. When the one is max the other is min. Maybe that's not what you meant.

How about you just post attempts using what you wrote?

 

[RoyceB]

So for c I decided at it's highest speed, Ee could equal 0 and for the other side I have Ee at it's highest and Ek = 0 so I ended up with.
Ee = Ek
½mv² = ½k(Δx)²
½(0.085)v² = ½(0.13)²(55)
v = 3.3m/s

b)

I made Ee + Ek = Ee' + Ek' and for the beginning I had it equal zero for both when it is not moving so I ended up with.
0 = Ee' + Ek'

Ee = -Ek
½mv² = ½k(Δx)²
½(0.085)v² = ½(55)(Δ0.074)²
v = 1.88m/s

Is that what you mean and it is similar for both of determining the energies.

 

[RoyceB]

Actually how about this for it.

a)
Ee = 0.46J

b)

Ee + Ek = Ee (max)
Ee (max) - Ee (at x = 0.074m) = Ek
½(55)(0.13)² - ½(55)(0.074)² = ½(0.085)(v²)
v = 2.78m/s

c)

Ek = Et
½mv² = Et
½(0.085)(v²) = 0.46J
v = 3.3m/s

 

[haruspex]

Actually how about this for it.

a)
Ee = 0.46J

b)

Ee + Ek = Ee (max)
Ee (max) - Ee (at x = 0.074m) = Ek
½(55)(0.13)² - ½(55)(0.074)² = ½(0.085)(v²)
v = 2.78m/s

c)

Ek = Et
½mv² = Et
½(0.085)(v²) = 0.46J
v = 3.3m/s

Looks good.