# Homework Help: Find the maximum velocity of the block

1. Dec 14, 2009

### PhilCam

1. The problem statement, all variables and given/known data

A small cart of mass .8 kg is attached to one end of a horizontal spring as shown in the figure. The spring constant is 600 N/m. When the spring is stretched by .05 m and the car is released from rest, the car oscillates back and forth. The friction is negligible. Ignore the mass of the spring.
A) Find the angular frequency of oscillations in HZ.
B) Find the maximum velocity of the block.
C) Find the total energy of the system at the point where acceleration is zero.

2. Relevant equations

Angular frequency = 2(3.14)F

3. The attempt at a solution
Pretty lost, if anyone could guide me to what formulas to use or what way to attack this problem I would be very grateful. Thanks

2. Dec 14, 2009

### mg0stisha

A) there is a much simpler equation for angular frequency using things you're given. they're found by solving SHO differential equations, did you do this in class?

B) the maximum velocity will occur when the acceleration equals...?

C) there's two types of energy involved. what are they?

3. Dec 14, 2009

### PhilCam

A) I have not found any other equation in my notes and to be honest, I'm not sure what "SHO" is.

B) When acceleration is zero, velocity will be maximum.

All the formulas I have for finding velocity include acceleration and by setting it to zero, it would be unsolvable.

I have formulas like F=kx but nothing that incorporates spring constant into velocity.

C) The two forces are potential and kinetic energy.
PE=1/2kx^2
PE= (.5)(600)(.05m)^2
PE= .75
is .75 Joules?

4. Dec 14, 2009

### ideasrule

SHO stands for simple harmonic motion. Have you seen the equation T=2pi*sqrt(m/k)?

5. Dec 14, 2009

### mg0stisha

Sorry, SHO is Simple Harmonic Oscillation. $$\omega=\sqrt{\frac{k}{m}}$$

What formulas for velocity do you have?

I think (but am not sure) that when the acceleration is 0, the spring would be at equilibrium (not stretched or compressed) therefore making the equation you used for energy equal to 0. Since they ask for max velocity (when acceleration is 0) and THEN ask for the energy when acceleration is equal to zero, you could use the normal kinetic energy equation, KE=(1/2)mv^2, using the maximum velocity you find in B.

Could anyone else confirm this?

6. Dec 14, 2009

### ideasrule

Confirmed. Alternatively, you could say that the kinetic energy is equal to the max potential energy, which is 0.5*k*x^2

7. Dec 15, 2009

### PhilCam

Thanks for the help guys.

I have figured out A and C but am still unsure on part B.

8. Dec 15, 2009

### mg0stisha

would conservation of energy work?
we now know that (1/2)mv^2=(1/2)kx^2, so try solving for v.

9. Dec 15, 2009

### PhilCam

Worked like a charm. thanks again.

10. Dec 15, 2009

No problem!