# Looking for an intuitive explanation for a mass-on-spring simple oscillator's frequency

Abdullah Almosalami
I just noticed something that is a little bit of a different perspective on a mass-on-spring (horizontal) simple (so undamped) oscillator's frequency and looking for some intuition on it.

There are many ways to derive that for a mass on a horizontal frictionless surface on a spring with spring constant k that the system's natural frequency is ##f = \frac {1} {2\pi} \sqrt {\frac k m}##. It makes intuitive sense. The larger the spring constant, the higher the restoring force and the faster that the mass will oscillate, and likewise the higher the mass, the more inertia it has, and the slower it responds resulting in lower frequency.

In my physics textbook, I was just doing some questions for practice and came across a question that asked "Use energy considerations to derive the oscillation frequency for a mass-on-spring oscillator ..." And ok. If I use the fact that the maximum potential energy due to the spring will be equal to the maximum kinetic energy of the mass (this will be the case for this system), I end up going through the following: $$PE_{max} = KE_{max} \Leftrightarrow \frac 1 2 kx_{max}^2 = \frac {1} {2} mv_{max}^2 \Leftrightarrow \frac {v_{max}} {x_{max}} = \sqrt {\frac k m}$$.

Now I know that ##\omega = \sqrt {\frac k m}## and dimensionally ##\frac {v_{max}} {x_{max}}## is indeed the unit of Hertz, but I'm not sure that I can immediately make the jump that this is indeed the oscillation frequency if I didn't already know what ##\omega## was from other derivations; so that's my question. What would make one conclude that this is indeed the oscillation frequency without using other derivations? I can also come to the same result by observing that if $$x(t) = x_{max} sin(\omega t) \Leftrightarrow v(t) = \frac {dx} {dt} = x_{max}\omega cos(\omega t) = v_{max} cos(\omega t)$$ then the ratio of the amplitude of the velocity to the amplitude of the displacement is indeed ##\omega##. But this is another derivation. I guess intuitively this result makes sense and provides an interesting perspective I hadn't thought of before: the sqrt of the maximum amplitude and the sqrt of the maximum speed are proportional to each other at a given frequency, and this makes for the known non-intuitive property of oscillators that ##\omega## is independent of amplitude. But yeah. My question is highlighted in bold.

Mentor
What would make one conclude that this is indeed the oscillation frequency without using other derivations?
Why wouldn't you use other derivations? Physicists don't forget everything that they have ever learned each time they approach a new problem.

Abdullah Almosalami
Why wouldn't you use other derivations? Physicists don't forget everything that they have ever learned each time they approach a new problem.

Fair point. I was just wondering if there is other insight that tells you that ##\frac {v_{max}} {x_{max}}## is the natural frequency without referring to other derivations.

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Fair point. I was just wondering if there is other insight that tells you that ##\frac {v_{max}} {x_{max}}## is the natural frequency without referring to other derivations.
SHM is probably a special case. The equation of motion is simple and compact, so perhaps there's nothing more to it than that.

• sophiecentaur and Abdullah Almosalami
You could also go down the following route if it wants energy considerations:$$E = T + V = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}kx^2 = \text{constant}$$ $$\dot{E} = m\dot{x} \ddot{x} + kx\dot{x} = 0$$ $$\ddot{x} = -\frac{k}{m}x$$Which is of course solved by ##x = A\cos{(\sqrt{\frac{k}{m}}t + \phi)}##, and you can identify ##\omega## in the normal way.

I guess you could also note that from the equations of motion you have ##\dot{x} = \pm \omega\sqrt{A^2 - x^2} \implies \frac{\dot{x}_{max}}{A} = \pm \omega##, which was the equation you derived.

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• • Leo Liu, vanhees71, Dale and 1 other person
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You can tie things together by looking at the situation from two different points of view. First consider the mass ##m## moving in the ##xy##-plane in a circle of radius ##A## with constant angular speed ##\omega.## The total energy of the mass is kinetic:$$KE=\frac{1}{2}mv^2=\frac{1}{2}m\omega^2A^2$$ and the position of the mass is given by $$x(t)=A\cos(\omega~t);~~y(t)=A\sin(\omega~t).$$ Now change your point of view and look at the ##xy##-plane edge-on with the ##y##-axis coming straight at you. You will see the mass moving back and forth along the ##x##-axis according to ##x(t)=A\cos(\omega~t).## However, its energy will still be ##KE=\frac{1}{2}m\omega^2A^2## because it should not depend on your angle of view. The first view allows you to see the total energy as ##KE=\frac{1}{2}m(v_x^2+v_y^2).## In the second view you don't see motion in the ##y##-direction and when the mass comes instantaneously at rest, it moves at maximum speed along the ##y##-axis. Knowing that the total energy is conserved, you conclude that the kinetic energy is stored sometimes in motion along one axis, sometimes along the other axis and sometimes is shared by motion along both. Sounds familiar?

Now let's say that you set up a spring-mass system on the ##xy##-plane such that ##\sqrt{k/m}## has the same value as the angular speed ##\omega## of the rotating mass. The rotating mass and the mass on the spring will have the same energy and will undergo identical edge-on view motions, assuming of course that you match their initial conditions. When a mass comes instantaneously to rest, (a) in the case of the rotating mass ##\frac{1}{2}mv_x^2## has become ##\frac{1}{2}mv_y^2~##; (b) in the case of the mass at the end of the spring ##\frac{1}{2}mv_{max}^2## has become ##\frac{1}{2}kA^2.##

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• vanhees71, Abdullah Almosalami, etotheipi and 2 others
Abdullah Almosalami
You could also go down the following route if it wants energy considerations:$$E = T + V = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}kx^2 = \text{constant}$$ $$\dot{E} = m\dot{x} \ddot{x} + kx\dot{x} = 0$$ $$\ddot{x} = -\frac{k}{m}x$$Which is of course solved by ##x = A\cos{(\sqrt{\frac{k}{m}}t + \phi)}##, and you can identify ##\omega## in the normal way.

I guess you could also note that from the equations of motion you have ##\dot{x} = \pm \omega\sqrt{A^2 - x^2} \implies \frac{\dot{x}_{max}}{A} = \pm \omega##, which was the equation you derived.

Right! I had seen that a while ago and forgot about it! Thank you! That is the insight I was looking for.

Abdullah Almosalami
However, its energy will still be ##KE=\frac{1}{2}m\omega^2A^2## because it should not depend on your angle of view.

Oh that's a nice way of looking at it! Another piece of insight I was looking for. Great!

Now let's say that you set up a spring-mass system on the ##xy##-plane such that ##\sqrt{k/m}## has the same value as the angular speed ##\omega## of the rotating mass. The rotating mass and the mass on the spring will have the same energy and will undergo identical edge-on view motions, assuming of course that you match their initial conditions. When a mass comes instantaneously to rest, (a) in the case of the rotating mass ##\frac{1}{2}mv_x^2## has become ##\frac{1}{2}mv_y^2~##; (b) in the case of the mass at the end of the spring ##\frac{1}{2}mv_{max}^2## has become ##\frac{1}{2}kA^2.##

Thanks!

Gold Member
Summary:: Looking for an intuitive explanation .
We have - in fact you have a perfectly good language (Maths) for describing (explaining even) what goes on. Why would one need to step backwards into 'intuition'? Imo, the way to work is to accept and use the maths of oscillations and let it become part of ones (so called) intuition. Remember, the terms that you expect in any answer you get on this thread will not be what you would have arrived on the Earth with. Your intuition was all learned at some stage.

• vanhees71 and weirdoguy
Abdullah Almosalami
Imo, the way to work is to accept and use the maths of oscillations and let it become part of ones (so called) intuition.

I have to disagree. Physics as I see it is not just applied math but is a combination of applying math and then stepping back, looking at the math, and making sense of it all from an intuitive (albeit perhaps mindbending at times) standpoint.

• • A.T., Delta2, sophiecentaur and 1 other person
weirdoguy
I have to disagree.

What you say works for classical non-relativistic mechanics for which we have some 'built in' intuition based on our everyday experience. It does not work with relativity, quantum physics and tons of other things for which we don't have any intuition to begin with. In those cases you have to build your intuition from maths. That's the way it is and it's better to accept it in the beginning of your journey, otherwise you'll get frustrated pretty quickly.

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We have - in fact you have a perfectly good language (Maths) for describing (explaining even) what goes on. Why would one need to step backwards into 'intuition'? Imo, the way to work is to accept and use the maths of oscillations and let it become part of ones (so called) intuition. Remember, the terms that you expect in any answer you get on this thread will not be what you would have arrived on the Earth with. Your intuition was all learned at some stage.
I cannot answer for the OP, but I can provide my own thinking on the "why". The preconceptions that students bring with them into the classroom are resistant to the correct mathematical development that we expect them to accept on faith. They have lived and operated on this Earth without difficulty all their life yet, when they enter a physics classroom, they are asked to view the world through math and ignore their everyday experiences.

Merely exposing students to the math associated with physics is insufficient to remove the preconceptions. For example, the preconception that "force implies motion" was examined [John Clement, Students' preconceptions in introductory Mechanics, AJP, 50, 66 (1982)] in what I believe is one of the earliest studies on the subject. Students were asked to consider a coin thrown vertically up and draw all the forces (neglecting air resistance) acting on it at a point where the coin is still rising. According to the study, 12% of the students got it right before taking a physics course, most of them showing a second force "from the hand" pointing up. After taking a physics course, the number of correct answers improved to only 28%.

My own conclusions from this is that special effort must be made to dislodge preconceptions based on intuition. Telling them about Newton's laws and showing them how to do implement them mathematically is not enough. That is why I use every opportunity to bridge intuition and math in the minds of people; one should modify, not reject, one's own intuition to the correct viewpoint which can only happen from within. Nevertheless, intuition is important in everyday life and its role should not be downplayed. Even the most adept physicist will prefer to rely on intuition to make a left turn from a parking lot into traffic than write down the math, solve the problem on a laptop and then proceed to make the turn.

• • Leo Liu and Delta2
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I have to disagree. Physics as I see it is not just applied math but is a combination of applying math and then stepping back, looking at the math, and making sense of it all from an intuitive (albeit perhaps mindbending at times) standpoint.
You seem to be assuming that one can do that on one's own. The process often fails (you can observe this on PF, often) unless a student can benefit from some dialogue with informed support. One may just not have had the appropriate experience to supply one with the necessary 'intuition'. The precise phrasing used in one particular textbook can be an insurmountable obstacle and it requires a lot of self confidence plus good maths ability to get some things sorted for oneself. Successful autodidacts are rare.

they are asked to view the world through math and ignore their everyday experiences.
Not in my classes with lower School Science. Once they get into Modern Physics they cannot choose to give their intuition the benefit of the doubt - if they want to learn about the subject. intuition (i.e. what they learned from their Grannies etc.) has to take second place. The Maths is just a part of the language they need to accept. You use the word "bridge" and that's the key.
One reason I have such a problem with simulations is that the same technology is used to show total nonsense from Hollywood and they cannot be expected to know what's Science and what's the notion of some film Director. Intuition can be completely wrong.
According to the study, 12% of the students got it right before taking a physics course, most of them showing a second force "from the hand" pointing up. After taking a physics course, the number of correct answers improved to only 28%.
My question is Which Physics course, delivered by whom? Intuition can be a tough nut to crack, of course. (My own included, sometimes)

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I don't know, what "intuitive" means. It's very personal for any individuum what he/she considers intuitive.

Though I personally find forces not intuitive at all, they are still something we are used to from everyday experience, and I'd say for this problem the most intuitive idea is to define the harmonic motion of a body as one, where the force is proportional to the displacement from a equilibrium position, which we take conveniently as the origin of our (1D) reference frame. Then the force is
$$F=-k x, \quad k>0.$$
Then Newton's equation of motion tells us that
$$m \ddot{x}=-k x,$$
from which immediately you get the general solution
$$x(t)=A \cos(\omega t) + B \sin (\omega t), \quad \omega=\sqrt{k/m},$$
with integration constants ##A## and ##B##.

I don't know, whether this is intuitive enough, but it's far better than using many words instead of a view clearly defined formulas. Math is your best friend in physics!

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I don't know, what "intuitive" means. It's very personal for any individuum what he/she considers intuitive.
Agreed. But people tend to much easier to say "I just know it", rather than "I learned it". Ask them what possible intuition a new born baby has. Then consider that some newly hatched migratory birds can travel thousands of miles and arrive at the right place - on their own. But, imo, they are not 'thinking about it'.

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Well, yes. That's the problem of humans being obviously "designed" being "multi-purpose" creatures, which is why they have to learn most of what they need. In physics you usually first have to unlearn what you think is intuitive and relearn the intuition by doing physics. It all starts with learning the only adequate language to talk about physics, which is math, no matter whether you tend more to the experimental or theoretical side!

• weirdoguy
I have to disagree. Physics as I see it is not just applied math but is a combination of applying math and then stepping back, looking at the math, and making sense of it all from an intuitive (albeit perhaps mindbending at times) standpoint.
I think if you do the math and then also try to apply intuition, that's perfectly valid. This is how you build a better intuition.

• vanhees71
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Not in my classes with lower School Science.
I am not familiar with lower School Science. My first (high school level) physics course relied on algebra and trig that I had already seen. What level math do you rely on to do your teaching? About how old are the students?

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Used to teach 11 to 18 years. A level used algebra and trig and a hint of diff calculus. Amazingly, it was not actually insisted that students didn’t do Maths courses. Things have changed a bit since I stopped tho’. Possibly for the better.

• kuruman
Gold Member
I never knew how they actually felt about the Maths. The three or four really bright students left Physics behind and did Medicine and Loadsa Moneymaking. Several went into teaching. Shame not to do something else before going into teaching, I always think. When you have to deliver a curriculum, that's the framework you tend to think in. So far and no further or they get scared. (Not the bright ones of course)

mfig
Here is some way to get at this that doesn't rely on having solved the system, but on some general considerations and a guess that the solution is sinusoidal. I am not sure this is what you are looking for, but hopefully it will contribute. $$PE_{max} = KE_{max} \Leftrightarrow \frac {x_{max}} {v_{max}} = \sqrt {\frac m k} = \gamma$$
and you want to know about this quantity γ. One thing to notice is that since in general $$speed = \frac{distance}{time}$$ then $$time = \frac{distance}{speed}$$ which is the form of your expression. So γ can be interpreted as the time it takes to travel the oscillation amplitude at the maximum oscillation speed. But since we know that the particle does not always travel at the maximum speed, the quantity is somewhat a mystery unless we can relate maximum speed to average speed. By definition, the average speed over one oscillation is the speed required to travel through one complete oscillation in one period:
$$v_{avg} = \frac{4 x_{max}}{\tau} = \frac{2 x_{max} \omega}{\pi} \rightarrow x_{max} = \frac{\pi v_{avg}}{2 \omega}$$
Call this equation 1.

Now if we make a guess that the motion is sinusoidal, then we can use the general result that the ratio of the maximum speed to the average speed is independent of frequency and amplitude. In other words:
$$v_{avg} = \frac{2 v_{max}}{\pi} \rightarrow v_{max} = \frac{\pi v_{avg}}{2}$$
Call this equation 2.

Finally, if we plug equations 1 and 2 into your result, we have:
$$\frac {x_{max}} {v_{max}} = \gamma = \frac{1}{\omega}$$
So, with your result, general considerations, and a guess that the motion is sinusoidal (but without solving the system to find the value of the frequency), we arrive at the conclusion that the time γ is the inverse frequency of oscillation.

I hope this helps in some way.