# Homework Help: Spring Mass System

1. Oct 21, 2011

### 626

Sorry for such a simple question. Unfortunately, this problem was originally written in Japanese and was just translated in English by a coworker. I don't know if there was a problem during the translation but we cannot produce the same answer as was shown in the Japanese answer key.

Problem:

A spring with a spring constant of K and with negligible mass was suspended, and vibrated with a small plate attached to its lower end. Then a cycle of To was obtained. When a weight with a mass of m1 was placed on the plate, the spring lowered by a length of a. Next, after the weight with a mass of m1 was removed, the spring was vibrated with another substance placed on the plate. In this case, if the obtained cycle was T, what is the mass m of this substance?

2. Oct 22, 2011

### Staff: Mentor

3. Oct 22, 2011

### 626

My Solution:

1. for the mass of m$_1{}$, elongation is a;

F = m$_1{}$g

but
F = ka

so:

k = m$_1{}$g/a

2. for the mass m which give a cycle of T

T = 2∏ √(m/k)

therefore:

m = m$_1{}$ [T$^2{}$/(4∏$^2{}$a/g)]

m = m$_1{}$ [(T$^2{}$-T$_o{}^2{}$)/(4∏$^2{}$a/g - T$_o{}^2{}$)]

Note: I pressume that since To was taken at no load (and spring mass is neglected) To can be considered 0. So substituting 0 for To for the expected answer, we can get the same relation, but still it is a bit confusing.

Can anyone provide an alternative solution that will give me the exact answer?

4. Oct 22, 2011

### Staff: Mentor

Do not assume that. Instead, assume there is some no load mass m0. When the mass m is added, the total mass is m0 + m.

5. Oct 22, 2011

### 626

Thanks but in that case result would be:

m = m1 [(T2-To2)/(4∏2a/g)]

still different from the expected answer of:

m = m1 [(T2-To2)/(4∏2a/g - To2)]

or am i missing something?

6. Oct 22, 2011

### Staff: Mentor

I don't think you're missing anything and I agree with your answer. Assuming you meant:
m = m1 [(T2-T02)/(4∏2a/g)]

7. Oct 22, 2011

### 626

sorry, just got it.

mo should be assigned in all equation including the first one (f=ma)

F=(mo + m1)g

Thanks for the help, I was able to get the exact solution using Doc Al's advice.

8. Oct 22, 2011

### Staff: Mentor

But the additional force when m1 is added is just m1g, not (m0 + m1)g. The length 'a' is the amount that the spring lowered when m1 was added.

Maybe something was lost in translation.

9. Oct 22, 2011

### 626

Yes perhaps.

But the simplest explanation that I can think of is that there was a mention of a small plate attached to the lower end of the spring. So assigning a mass of mo to this plate makes sense.

"When a weight with a mass of m1 was placed on the plate..." so the deflection (a) is really caused by the weight of the plate mo and the added mass m1.

A little tricky but at least we can get the answer that we are looking for.

10. Oct 22, 2011

### Staff: Mentor

Right.
Deflection from what? From its original position before the mass m1 was added, which already included any stretch from the m0 mass.

That's the simplest interpretation of "When a weight with a mass of m1 was placed on the plate, the spring lowered by a length of a."
I'd say badly worded. But no matter.

11. Oct 22, 2011

### 626

By the way, this question was part of a ME competition questionnaires held in our office in Tokyo. A bit tricky in my own opinion.

Another tricky question:

Problem:

A 2200kg rotor is connected to a 150kg pinion shaft by a coupling with an axial spring stiffness of 7.06x10^6N/m. What is the frequency of the maximum axial vibration?

i used the 150kg mass in my solution:

f = sqrt(k/m)/2pi which gave me 34.5Hz

i guess my solution was incorrect but i'd like to know the right solution.

12. Oct 22, 2011

### 626

Yes you're absolutely right. So the best translation would be:

"When a weight with a mass of m1 was placed on the plate, the spring lowered by a total length of a."

13. Oct 22, 2011

### Staff: Mentor

Right. Compared to its unstretched length.

14. Oct 22, 2011

### 626

By the way, this question was part of a ME competition questionnaires held in our office in Tokyo. A bit tricky in my own opinion.

Another tricky question:

Problem:

A 2200kg rotor is connected to a 150kg pinion shaft by a coupling with an axial spring stiffness of 7.06x10^6N/m. What is the frequency of the maximum axial vibration?

i used the 150kg mass in my solution:

f = sqrt(k/m)/2pi which gave me 34.5Hz

i guess my solution was incorrect but i'd like to know the right solution.

15. Oct 22, 2011

### Staff: Mentor

I'm not sure I know what a rotor or pinion shaft is.

Nonetheless, you treated the 2200 kg mass as if it were infinite. Instead, consider the vibration about the center of mass, which effectively shortens the spring. That will account for the greater frequency.

16. Oct 22, 2011

### 626

:) sorry but what do you mean "center of mass". unfortunately, the values stated are the only variables given in the problem. so how do I calculate/find the center of mass? :)

btw, i think the rotor and pinion can just be considered as two masses joined together by a spring and subjected to axial rotation. but no RPM was given.

17. Oct 22, 2011

### Staff: Mentor

Imagine the spring has length L. You have two masses, one on each end of the spring. The entire system will oscillate about the center of mass. Where is it?

Once you find the center of mass (in terms of L), then you can treat each mass as being on the end of a shorter spring that is fixed at the center of mass. Find the spring constant of that shorter spring (for the smaller mass, say).
I suspect an axial vibration, not a rotation. (Movement is parallel to the axis.)