Finding Initial Block Position and Maximum Speed of Oscillation

[NY152]

1. The problem statement, all variables and given/known
On a very smooth horizontal table, a block of mass m=0.75 kg is attached to an ideal spring with a spring constant k= 242 N/m. The origin of the horizontal coordinate (x=0) is set at the equilibrium position of the block. The block is initially held at a negative position where it keeps the spring compressed. Then the block is released, and moves through position x1=0.105m with a speed v1=1.42 m/s.

a) What was the initial position of the block, x0?

b)What is the maximum speed of oscillation, vmax?

Homework Equations

omega = sqrt(k/m
F=-kx
Vmax=A*omega
x=Acos(2pi*f*t)

The Attempt at a Solution

I thought about using conservation of energy by doing KE+PE=W
W=Fs and KE=1/2mv^2 but I don't think this would give me displacement in terms of a negative/before equilibrium position

 

[gneill]

Hi NY152,

Welcome to Physics Forums.

I thought about using conservation of energy by doing KE+PE=W
W=Fs and KE=1/2mv^2 but I don't think this would give me displacement in terms of a negative/before equilibrium position

If you take a close look at the formulas for the components of the energy, both involve the squaring of a term. A negative value squared yields the same amount as a positive value of the same magnitude.

 

[NY152]

Hi NY152,

Welcome to Physics Forums.
If you take a close look at the formulas for the components of the energy, both involve the squaring of a term. A negative value squared yields the same amount as a positive value of the same magnitude.

Thanks for the input! Do you think I'm on the right track then in terms of those formulas or am I missing something??

 

[gneill]

You're on the right track. You're just missing stating the formula for the PE of a spring.

 

[NY152]

You're on the right track. You're just missing stating the formula for the PE of a spring.

I only have one attempt left, so I'm just going to post my work here in the hopes that someone might see if I'm doing it right or wrong.

So I did: KE+PE=W W=Fs
1/2mv^2+1/2kx^2=(-kx)(s)
and found s to be =-0.0822

 

[gneill]

I only have one attempt left, so I'm just going to post my work here in the hopes that someone might see if I'm doing it right or wrong.

So I did: KE+PE=W W=Fs
1/2mv^2+1/2kx^2=(-kx)(s)
and found s to be =-0.0822

No, the sum of the KE and PE is a constant, the total energy of the system. Better to write:

E = KE + PE

In an isolated system the KE and PE can trade energy back and forth, but their sum is always a constant.

You are given a particular data point with both displacement and velocity. So you can find the value of E which will hold for every point throughout the cycles.

 

[NY152]

No, the sum of the KE and PE is a constant, the total energy of the system. Better to write:

E = KE + PE

In an isolated system the KE and PE can trade energy back and forth, but their sum is always a constant.

You are given a particular data point with both displacement and velocity. So you can find the value of E which will hold for every point throughout the cycles.

No, the sum of the KE and PE is a constant, the total energy of the system. Better to write:

E = KE + PE

In an isolated system the KE and PE can trade energy back and forth, but their sum is always a constant.

You are given a particular data point with both displacement and velocity. So you can find the value of E which will hold for every point throughout the cycles.

Sorry for the late reply, so I solved for E and got 2.09, now do I just set that equal to PE + KE when the spring is retracted? In which case would velocity be zero? If so I get x=-0.131

 

[gneill]

Sorry for the late reply, so I solved for E and got 2.09, now do I just set that equal to PE + KE when the spring is retracted? In which case would velocity be zero? If so I get x=-0.131

Yup. Good.

 

[NY152]

Sorry for the late reply, so I solved for E and got 2.09, now do I just set that equal to PE + KE when the spring is retracted? In which case would velocity be zero? If so I get x=-0.131

Alright well it wasn't -0.131, but I got the max velocity right by setting x=0 so I'm not sure where I went wrong but thanks for the help!

 

[gneill]

Alright well it wasn't -0.131, but I got the max velocity right by setting x=0 so I'm not sure where I went wrong but thanks for the help!

Your initial position should be correct. Did you remember to include the units?