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

A block rests on a frictionless horizontal surface and is attached to a spring. When set into simple harmonic motion the block oscilates back and forth with an angular frequency of 7.0 rad/sec. The drawing indicated the position of the block when the spring is unstrained x=0. The drawing also show a small bottle located .080m to the right of this position. The block is pulled to the right, stretching the spring by .050m, and is then thrown to the left. In order for the block to knock the bottle over, it must be thrown with a speed exceeding vo (initial velocity). Ignoring the width of the block find vo.

2. Relevant equations

I'm not sure need serious help.

3. The attempt at a solution

w=2pi/x w=.898
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 Blog Entries: 5 Recognitions: Homework Help Science Advisor Welcome to PF. Here is the general approach: You will need to find the equation of motion. For this, use F = ma. What is the force? What is the acceleration in terms of the displacement x(t)? This will give you a second order differential equation. What are the boundary conditions? Then find the maximum and solve it for v0. Just start and show us what you got and where you get stuck, so we can provide more detailed help.
 Thank you CompuChip. I don't understand how i'm supposed to find the force since the mass of the block is not given and since I dont know the force I'm not sure how to find the acceleration, and what are boundary conditions? We haven't learned about those in class yet. I mean the only thing the problem provides is the distance the spring travels, and the period while oscilating. Can you jump start me on what equations i need to solve this. Thank you

Blog Entries: 5
Recognitions:
Homework Help

You will need the mass, so let's call it m. The force that a spring exerts is $F = - k x$, where k is a constant (it's a property of the string, indicating how easily it stretches and compresses) and x is the displacement from the equilibrium position of the string (if the spring is in equilibrium, there is no force; if you stretch it in one direction, the spring will exert a force the other way to try and restore it to x = 0). This is called Hooke's law.
The acceleration is $a(t) = x''(t)$, the second derivative with respect to time. So if you use Newton's law, you get
$$- k x(t) = m x''(t)$$