# Compression question

1. Nov 3, 2007

### LostTexan07

The springs of a 1100 kg car compress 6.0 mm when its 68 kg driver gets into the driver's seat. If the car goes over a bump, what will be the frequency of vibrations?

I tried to use the weight of the man (68 kg) and the compression distance (6 mm) to find the spring constant. I then tried to use the spring constant to find the period.

F = kx
68 = 6k
k = 11.33

T = 2(pi)(sq.rt 1100/11.33)

But I didn't get the correct answer.

2. Nov 3, 2007

### learningphysics

When you calculate k, first convert mm to m. 6.0mm = 6.0*10^-3m

When you calculate T, use the mass of the man = 68kg. not the mass of the car.

3. Nov 5, 2007

### dynamicsolo

Also, the driver's mass is 68 kg., but you need the added weight of the driver applied to the springs. The spring constant, k, needs to be expressed in SI units, which are Newtons/m, which is why learningphysics corrected your 6 mm entry.

For the period of oscillation, you will need the combined mass of car and driver. (I'll say that the answer is in the neighborhood of 0.6 second.)

Last edited: Nov 5, 2007
4. Nov 5, 2007

### learningphysics

Yes, sorry you need to use:

F = kx

mg = kx (with m = 68kg) to find k...

I didn't notice the g was missing.

dynamicsolo, I'm not getting 0.6 for the period like that... are you sure about using the combined mass of the car and the person?

5. Nov 5, 2007

### dynamicsolo

You do use the total mass (1168 kg.) now resting on the springs for this "oscillator". The figure I gave for the period is approximate because I didn't want to just provide the answer. [k should be around 111,000 N/m and sqrt(m/k) is roughly 0.1 sec.]

6. Nov 5, 2007

### learningphysics

Oh, I see my problem. I was thinking the springs described were "in" the driver's seat... sorry about that.

yes, I get around 0.6. I must have been plugging in something wrong.