# Simple Harmonic Motion and maximum velocity

1. Mar 8, 2008

### kiro626

[SOLVED] Simple Harmonic Motion

I'm really having a hard time solving these two problems.

1. Objects of equal mass are oscillating up and down in simple harmonic motion on two different vertical springs. The spring constant of spring 1 is 174 N/m. The motion of the object on spring 1 has twice the amplitude as the motion of the object on spring 2. The magnitude of the maximum velocity is the same in each case. Find the spring constant of spring 2.

2. Relevant equations
the relevant equations are those for SHM. F=kx and f=-kx.

3. The attempt at a solution
I tried the formulas but I still can't find the answer. The answer is supposed to be 696N/m

2nd problem:

A small ball is attached to one end of a spring that has an unstrained length of 0.200 m. The spring is held by the other end, and the ball is whirled around in a horizontal circle at a speed of 3.00 m/s. The spring remains nearly parallel to the ground during the motion and is observed to stretch by 0.010 m. By how much would the spring stretch if it were attached to the ceiling and the ball allowed to hang straight down, motionless?

2. Relevant equations
same as the one above

3. The attempt at a solution
Again with this one, I tried solving it using the same formula but I can't figure how to use the velocity since its in m/s rather than rad/s for uniform circular motion. The answer is supposed to be 2.29 x 10^-3

Any help would be gladly appreciated

Last edited: Mar 8, 2008
2. Mar 8, 2008

### Kurdt

Staff Emeritus
This is quite a neat problem. The position of the spring is given by $y = A sin(\omega t)$ where $A$ is the amplitude and $\omega = \sqrt{\frac{k}{m}}$. You need to work out what the velocity will be and then set them equal. REmember the max velocity will be at the equilibrium point.

3. Mar 8, 2008

### kiro626

I already solved the first problem but I still can't figure the second one out. I tried equating the constant of spring since they would be equal but it give me 2.18x10^-3 so its a little off

Last edited: Mar 8, 2008
4. Mar 8, 2008

### Kurdt

Staff Emeritus
WEll if you show your full work it would help me. I suspect you're using the wrong value for r.

5. Mar 8, 2008

### kiro626

uhmm... for the second problem i used the equation:

ma = kx

wherein a is centripetal acceleration
i used 0.2 as the r and then 0.01 as the x to get the k then i equated the k from the second equation that i made which is:

mg = kx(sub2)

I'm not really sure about it

Last edited: Mar 8, 2008
6. Mar 8, 2008

### physixguru

it shall do a lot of good if ya tell us why are ya not sure?

7. Mar 8, 2008

### Kurdt

Staff Emeritus
I'd have used 0.21 as r.

8. Mar 8, 2008

### kiro626

thanks I finally got the answer. I finally understood it ^^ thanks very much.