# Period of oscillation of a suspended mass

My initial approach was using the formula $F = - kx$ and acquiring the spring constant and later inputting it in $T = 2\pi \sqrt \frac{m}{k}$ to get the period.

Although, my friend contacted me and told me that he talked to the teacher about it who said the spring constant can not be negative. That is what's confusing me. Other than that, the second question's answer. If we increase the mass by $1000$, the period will be multiplied by root of $1000$. But there will still be oscillations, so I don't get why the opposite could happen.

Consider a spring suspended from a firm support on the ceiling. A mass is then suspended on the other end of the spring. When the spring is stretched by a pair of 2.4 N forces, its length is found to increase by 62 mm.

a. Predict the Period of oscillation when an object of mass 0.32 kg is suspended on the spring.

b. Is it possible to increase the mass by 1000 times and still have oscillations? Why? State all the assumptions that you may have made while analyzing the spring-mass system in this question.

The formula $F=-kx$ comes from the more proper vector form $\vec{F}=-k\vec{x}$ and tells us the relationship between the force which a spring exerts on things it interacts with and the direction and amount of spring distortion ($\vec{x}$).

If the coils of the spring are in equilibrium, then it also tells us the magnitude of the force being exerted on each end of the spring.

The negative sign tells us that the direction of the force the spring exerts is opposite the direction of the distortion: if the spring is compressed/stretched from its self-equilibrium, it will exert a force opposite that compression/stretch. Therefore, in determining $k$ from the given information you should not make it negative: the sign is for vector direction purposes.

Regarding the mass increase ($1000\times$), I agree with you. Short of a mechanical failure or lack of room to operate, the system should still oscillate. There are large structures (buildings, ships) which oscillate.

• "Short of a mechanical failure or lack of room to operate" You can actually do the math on that with the values given in part A. Suffice to say, unless this is an extremely high ceiling, you might run into some issues. – JMac Sep 11 '17 at 20:34
• @JMac Think about the sizes of ocean freighters, building cranes and strip mining equipment. There is some really huge equipment out there. As Floris calculated, an 18 m spring is not beyond reality in some of these machines. – Bill N Sep 11 '17 at 22:06

Bill N helped explain the problem of the sign of the spring constant - but didn't address the issue of oscillation if you increase the mass by 1000x. Then he pointed out a problem with my math - calculations now updated with a tip of the hat to him!

If the spring extends by 62 mm with a force of 4.8 N, then it would have to extend (within the linear regime) to a length of (320*9.8/4.8)*0.062 = 40 m when loaded with a mass of 320 kg. While that is conceivable, it would not be so in the room where the experiment is done. Instead, after the spring stretched about 2.5 m, the mass would "lie on the floor, motionless".

Assuming there is enough space (you are doing this with a mass suspended from a high beam at the top of a tall staircase, perhaps), you still have the question "will the spring allow you to stretch it this much?". And that means that the total length of the coiled spring must be many times greater than 40 m - otherwise, it will simply no longer be a coiled spring.

Let's assume the total length of coiled wire is 100 x longer - then the total length of coil is about 4 km. Wrapping this in a coil that is about 50 cm diameter (1 turns = 1.5 m), you would need about 2500 turns.

The general equation for a spring constant is

$$k = \frac{Gd^4}{8 D^3 N}$$

Where G = shear modulus, d=wire diameter, D = coil diameter, N = number of turns.

We can then estimate the thickness of the wire we need for this spring:

$$d = \sqrt[4]{\frac{8 k D^3 N}{G}}$$

For steel, G = 75 GPa. For a 2500 turn coil of 50 cm diameter to have a spring constant of 4.8 N / 62 mm (774 N/m), it would have to have to be made with a steel wire that is 4 cm thick (!) - and the ratio of d/D would be about 1/12, which is at the low end of what is considered OK for springs. With 2500 turns, the unstretched length of this spring is a little over 100 m. It is not inconceivable that such a spring would stretch by 40 m when loaded with a 320 kg object, and that it would still oscillate. You'd need a pretty big staircase to accommodate the experiment; and it's clearly not what the question had in mind...

Usually, springs that can carry heavy loads will have greater stiffness, less deflection. Because we're trying to make a spring here that has a fairly low spring constant, yet make it capable of carrying a large load, we have to make it very very big.

• do you mean 40 meters instead of 4 (several places)? Yeah, that kind of spring may seem ridiculous, but my son helps build dredging equipment, and that stuff is huge. I wouldn't doubt that something like that actually exists. My real point is that large massive structures do oscillate. – Bill N Sep 11 '17 at 22:02
• 62 mm = 0.062 m, not 0.0062 m – Bill N Sep 11 '17 at 22:09
• @BillN - no, I think I do mean 4 meters. 6 mm with 4.8 N... so a weight of 3200 N would be less than 6 m. At one point I was trying to make a spring with a larger diameter - but as you do that, dimensions quickly get (much) worse. Not sure what the correct ratio of wire diameter to spring diameter would be. If you want to make d/D small, then D must be enormous. I agree "it would be possible". It would be one monster spring though. – Floris Sep 11 '17 at 22:11
• Ah - I have 6.2 mm in my equations... thanks. – Floris Sep 11 '17 at 22:11
• Though I thought about it some more, and even that is ambiguous for the springs motion. This is just another great textbook question. So I don't know now either. Either way, 40 m is quite a bit. – JMac Sep 11 '17 at 22:39

"spring constant cannopt be negative": in F=kx, F is the force required to be applied to the spring to stretch it. It has the same direction as x. (it is not the force applied by the spring.)

Question b) Can a spring extend for-ever? Are there any limits defined in this question?

There's a damping factor. Your damping factor has to be less than that in order to have oscillations. You can calculate that damping factor: 2 sqrt (m*k)

• Damping factor has to be less than what? Also, considering this only mentions a spring and shows Hookes law, I don't think it's safe to assume they wanted damping. I honestly don't think that is what the question was looking for. – JMac Sep 11 '17 at 17:33