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Simple Harmonic Motion, memory device

by oneplusone
Tags: device, harmonic, memory, motion, simple
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oneplusone
#1
Jan15-14, 04:26 PM
P: 127
What is a good way to memorize that ## \omega = \sqrt{\dfrac{k}{m}} ## ?
I always confuse it with: ## T = 2\pi \sqrt{\dfrac{m}{k}}## , and can never tell them apart. (i guess part of it is that I'm not too familiar with it yet)
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#2
Jan15-14, 04:54 PM
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Do latex with [itex]\omega = \sqrt{\dfrac{k}{m}}[/itex].
tiny-tim
#3
Jan15-14, 04:59 PM
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hi oneplusone!

(you need to put two #s either side of your latex )
Quote Quote by oneplusone View Post
What is a good way to memorize that ##\omega = \sqrt{\dfrac{k}{m}} ## ?
I always confuse it with: ##T = 2\pi \sqrt{\dfrac{m}{k}}## , and can never tell them apart. (i guess part of it is that I'm not too familiar with it yet)
easy-peasy
mx'' = -kx, so m/k = -x/x'' = time squared

T is time, so ##T = 2\pi \sqrt{\dfrac{m}{k}}##

thegreenlaser
#4
Jan15-14, 05:31 PM
P: 461
Simple Harmonic Motion, memory device

Quote Quote by oneplusone View Post
What is a good way to memorize that \[ \omega = \sqrt{\dfrac{k}{m}} \] ?
I always confuse it with: \( T = 2\pi \sqrt{\dfrac{m}{k}}\) , and can never tell them apart. (i guess part of it is that I'm not too familiar with it yet)
When you have two formulas and you know one of them is correct, I find the best way is often just to reason it out. Which formula makes the most sense?

Suppose [itex]\omega = \sqrt{k/m}[/itex]. That formula tells me that the oscillations are faster when k (spring stiffness) is increased and/or m (mass) is decreased.

Suppose [itex]\omega = 2\pi \sqrt{m/k}[/itex]. That formula tells me the opposite: the oscillations are faster when k is decreased and/or m is increased.

Now, which one of those makes sense? If you keep loosening the spring, what would you expect to happen? If you think about it, it should be clear that you would expect to see big sweeping oscillations which take a lot of time, and thus you would have fewer oscillations per second (i.e. lower frequency). So the first formula has to be the right one because the second formula is telling you that a weaker spring will have faster oscillations, which doesn't make sense when you think about it. The first one agrees with what makes sense.

Another trick would be to use units. Personally, I know the units of m (kg), ω (rad/s), and T (s), but I don't know the units of k. However, I do remember Hooke's law (F = -kx), so I can easily figure out that k is measured in kg/s2. From that, I can figure out that sqrt(m/k) would have units of seconds, which means ω = 2π*sqrt(m/k) is nonsense since I have frequency on one side and seconds on the other. ω = sqrt(k/m), on the other hand, works out unit-wise (there's some funny business with radians here, but at least you can see that ω = 2π*sqrt(m/k) is definitely wrong).

Maybe you were looking for some sort of mnemonic, but the methods I've described are pretty useful for all sorts of situations like this. Basically, you use what you know to fill in the gaps of stuff you don't know. I don't have an amazing memory, personally, and I quite often find myself in the situation where I only roughly remember a formula. Usually I remember what quantities are involved, but I can't remember which one's in the numerator or denominator, or I can't remember whether it's positive or negative. However, I've just learned to very quickly figure out what the correct formula is by using what I do know (units, other formulas, conceptual understanding of the physics) to see which "version" of the formula makes sense. You can almost always figure out the correct formula that way if you at least have a decent understanding of the subject.
Badfish97
#5
Jan18-14, 01:17 PM
P: 13
Quote Quote by oneplusone View Post
What is a good way to memorize that ## \omega = \sqrt{\dfrac{k}{m}} ## ?
I always confuse it with: ## T = 2\pi \sqrt{\dfrac{m}{k}}## , and can never tell them apart. (i guess part of it is that I'm not too familiar with it yet)
Its real simple ω=2πf [f(frequency)=1/T]
which implies, T=2π/ω
since you already know ω=√k/m,
just substitute and you'll get the expression for T.
Just remember ω=√k/m & ω=2π/T...look up the complete derivations if your still having trouble.
sophiecentaur
#6
Jan18-14, 02:40 PM
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If you are prepared to put up with countless ∏s turning up in your calculations then you never need to deal with ω. Just remember T = 1/f (which makes sense).
'Big boys' use ω because the Maths behaves much better when you do.
I wouldn't mind betting that. after you have gone through this whole thread, you will not find it a problem, in any case. lol. (Learning by doing)


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