# SMH Vertical Spring problem

1. Nov 5, 2013

### rmiller70015

1. The problem statement, all variables and given/known data
A 2.0kg pumpkin oscillates from a vertically hanging light spring once ever 0.65s. Write down the equation giving the pumpkin's position as a function of time, assuming it started by being compressed 18cm from equilibrium. How long will it take to get to the equilibrium position for the first time?

2. Relevant equations
y=Acos(2πt/T)

3. The attempt at a solution
The equation I got was y=0.18mcos(2πt/0.65s), which is correct according to the answer in the back of the book, but I am having trouble finding the time.

What I did to find the time was to consider the object at the beginning of the motion, that is t=0 and y=.18m. So to solve for t, I divided by the 0.18m or A :
y/A=cos(2πt/T)

From here I used the arccos function to get the t variable out of the cosine function and got:
arccos(y/A)=2πt/T

Then I multiplied by T and divide by 2π to get
Tarccos(y/A)/2π=t

But the answer I get from this is not correct, something like 0.33s when it should be something like 0.16s.

2. Nov 5, 2013

### Staff: Mentor

What part of a full cycle is represented by the trip from release point to equilibrium?

3. Nov 5, 2013

### rmiller70015

On the graph of the function f(x)=0.18cos(2πx/0.65) it would be between x=0 and x≈0.16, I got that by using google to graph the function and found the first y intersect, which should be the equilibrium point for the motion, but I'm still unclear on how to do it mathematically, I guess I could set the equation equal to zero and solve for it, but I end up getting t=0. I guess it is essentially the first quarter of the motion though.

4. Nov 5, 2013

### Staff: Mentor

You're probably getting zero for t because you're solving for when the pumpkin is at 18 cm, it's initial position. That's not where the equilibrium position is. What is y at the equilibrium position?

Not that you can solve this problem without using the equation if you will consider what part of a full cycle the trip from release to equilibrium position represents.

5. Nov 5, 2013

### rmiller70015

Yeah I realized that by replacing 2π with π/2 and you get the correct answer using the cos arccos method.

But aren't I measuring it from the equilibrium position, since 18cm in the equation is the amplitude (A) in the function Acos(2πt/T)=y=0, where y would be the position at equilibrium since (from the original problem, "...assuming it started by being compressed 18cm from equilibrium.")

6. Nov 5, 2013

### Staff: Mentor

Sure, the position y is with respect to the equilibrium position. [STRIKE]But I don't see what point you're trying to make.

What is the value of y at the equilibrium position? So what equation do you want to solve to find t?[/STRIKE]

EDIT: I just re-read your reply and realized that you have the correct notion; sorry about that.

Yes, for the equilibrium position you have Acos(2πt/T)=0. Divide though by A and you have cos(2πt/T)=0.
What is the first angle θ for which cos(θ) is zero?

Last edited: Nov 5, 2013
7. Nov 5, 2013

### rmiller70015

Thanks for your help, after realizing what you meant by what portion of the motion is it to equilibrium, I divided 0.65 by 4 and then felt really stupid.