Simple Harmonic Motion, memory device

AI Thread Summary
To memorize the formula ##\omega = \sqrt{\dfrac{k}{m}}##, understanding the relationships between the variables can be helpful. The formula indicates that oscillation frequency increases with higher spring stiffness (k) and lower mass (m). In contrast, the formula for period ##T = 2\pi \sqrt{\dfrac{m}{k}}## suggests that a weaker spring leads to slower oscillations, which is counterintuitive. Using unit analysis can also clarify which formula is correct, as the units must align appropriately. Developing a conceptual grasp of the physics behind these formulas enhances retention and understanding.
oneplusone
Messages
127
Reaction score
2
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)
 
Last edited:
Physics news on Phys.org
Do latex with \omega = \sqrt{\dfrac{k}{m}}.
 
hi oneplusone! :smile:

(you need to put two #s either side of your latex :wink:)
oneplusone said:
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}}## :wink:
 
oneplusone said:
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 \omega = \sqrt{k/m}. That formula tells me that the oscillations are faster when k (spring stiffness) is increased and/or m (mass) is decreased.

Suppose \omega = 2\pi \sqrt{m/k}. 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.
 
  • Like
Likes 1 person
oneplusone said:
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.
 
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)
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Hello everyone, Consider the problem in which a car is told to travel at 30 km/h for L kilometers and then at 60 km/h for another L kilometers. Next, you are asked to determine the average speed. My question is: although we know that the average speed in this case is the harmonic mean of the two speeds, is it also possible to state that the average speed over this 2L-kilometer stretch can be obtained as a weighted average of the two speeds? Best regards, DaTario
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Back
Top