# How to derive an equation for a horizontal frictionless harmonic oscillator that has experienced sudden hit by a hammer?

Here (and here), I saw an equation for a horizontal frictionless harmonic oscillator (a mass on a spring) that was suddenty hit by a hummer (i.e. the duration of the hit goes to zero). The equation looks like this (neglecting the coefficients): $$x'' + x = \delta_{\pi}(t)$$ My question is how to derive it.

Some clarification of what I'm specifically interested in.

I know how to derive the left hand side but I want to know how do we end up having Dirac delta on the right hand side.

Normally, we would draw a free body diagram for the mass and write down the second Newton's law in projections on the horizontal axis $$x$$. Thus, for the moment of the hammer hit we have: $$ma = F_{spring} - F_{hammer}$$ $$x''-\frac{k}{m} x = - \frac{1}{m} F_{hammer}$$ At this point, we've derived the left hand side. How do we convert the right hand side to a Dirac delta containing function?

As I understand, the second Newton's law that I have written works only for the moment of the hammer hit. That's because $$F_{hammer}$$ exists only at the moment of the hammer hit - say - at $$t=0$$. In other words, $$F_{hammer} (t)$$ looks like this: In order to use our equation for all the values on the $$t$$ axis, we have to "expand" $$F_{hammer} (t)$$ to the entire $$t$$ axis. I believe that Dirac delta has something to do with that. I.e. Dirac delta converts our plot to the following one: I'm very curious about the essence of the mathematical trick that expands the function $$F_{hammer}$$ which exists only at one point to the entire $$t$$ axis.

• Do you think it should be something different from a delta function? Or you simply have never encountered delta functions? Why do you think it's a "mathematical trick"? Aug 7, 2021 at 19:15
• The hammer gives an impulse (the technical term for a change in momentum) to the system. You find the initial velocity by conservation of momentum, exactly the same way as any problem about "collisions" in Newtonian mechanics. (And you don't need to know anything about delta functions to do this). Aug 7, 2021 at 19:58
• Side note: The linked video uses $\delta_\pi(t)$ which is kind of a strange notation. Usually this is written as $\delta(t-\pi)$. Aug 7, 2021 at 21:04
• @BillN , I think that it should be something different from a delta function because I get $F_{hammer}$ instead of a delta function when I try to derive the equation. $F_{hammer}$ and Dirac delta are different things. The question is how I convert $F_{hammer}$ to the Dirac delta. I have never encountered Dirac delta before, that's true. I suggested that's a mathematical trick because I decided that we expand a function to the entire $R^3$ when we multiply it by Dirac delta. Otherwise, I don' understand how multiplying by Dirac delta changes a function. Aug 7, 2021 at 22:47

You seem to be new to the concept of the $$\delta$$-function. Therefore I will try to motivate this now.

Let's hit the oscillator body with a hammer made of soft rubber. The force-time graph will look like this: The force is non-zero for a short time, while the hammer is in touch with the oscillator body. The rubber is compressed first, and then bounces back to its original shape. The momentum transferred from the hammer to the oscillator body can be calculated by $$\int_{-\infty}^{+\infty} F_\text{hammer}(t)\ dt = P$$

Now let's hit it with a hammer made of hard steel. Then the force-time graph will look like this: The situation is similar to the situation above, except that the hitting time is even shorter, and the force during this time is even larger. But the transferred momentum is still the same: $$\int_{-\infty}^{+\infty} F_\text{hammer}(t)\ dt = P$$

Now let's use an infinitely hard hammer (of course such a thing doesn't really exist, but let's do it anyway). The hitting time is zero now, and the hitting force during this time is infinitely high. We write the force now using the Dirac delta function as: $$F_\text{hammer}(t)=P\ \delta(t)$$ Once again the transferred momentum is the same (now by definition of the $$\delta$$-function): $$\int_{-\infty}^{+\infty} F_\text{hammer}(t)\ dt = \int_{-\infty}^{+\infty} P\ \delta(t)\ dt = P\ \underbrace{\int_{-\infty}^{+\infty}\delta(t)\ dt}_{=1} = P$$

The $$\delta$$-function is just a mathematical idealization (originally invented by physicists) for modeling infinitely narrow and high spikes.

• Great answer tf. I guess the two hammers same only if elastic collision w block. (Or both inelastic and same mass - same mass since it is now on the block lol). Or same losses in both cases. But not ness for this q. Aug 7, 2021 at 20:48
• Thank you @Thomas Fritsch! The answer is great indeed. Can we clarify one thing, please, and I will mark your response as the solution? In order to represent $F_{hammer}$ on the right hand side of my equation through Dirac delta, I substitute $P \delta (t)$ for it. Then I say that $P=mx'$ and I get a differential equation with respect to x. Am I right? Aug 7, 2021 at 22:36
• @IvanNepomnyashchikh The $P$ in $F_{hammer}(t)=P\delta(t)$ is just a constant giving the momentum transferred by the hitting hammer. Do not confuse it with $p(t)=m\dot{x}(t)$ which is the time-dependent momentum of the oscillator body. Aug 7, 2021 at 22:52
• Thank you @ThomasFritsch! That was exactly my concern. But the rest is right? We substitute $P \delta (t)$ for $F_{hammer}$ and so on - that's correct, yes? Aug 7, 2021 at 22:54
• @IvanNepomnyashchikh Yes, it is only for that. Aug 7, 2021 at 22:56

The only thing I’ll add to Thomas’ great answer is the terminology of an “impulse”, sometimes called “impulse force”.

If you say “And then there is an impulse on the object.”, people will get that it is a very short-term, large force, such as an impact, and we don’t know or care about the specific dynamics of that force, just its net effect, which is an instant transfer of momentum. (Impulse is usually an impact but not always).

That’s because:

1. It’s of short enough duration that we can assume instantaneous, nothing else in our system will change enough to matter.

2. it’s hard to characterize the force/time dynamics, hard to measure or calculate whereas the net result is not hard to determine.

I’d say Impulse is a very high force for a short time characterized not by those force-time dynamics but by the momentum (or equivalently energy) transferred,

(Where technically that net effect is from integrating the force over some short time.)

• Great discussion. Let me add my little contribution to this regarding country wise differences. In Russia, we call "impulse" what is called "momentum" in the US. We use it as the official term. And we call $Ft$ "impulse of force". At best, we can call $mv$ "quantity of motion" which is called this way in the Newton's "The mathematical principles of natural philosophy". When I first came to the US it was really hard for me to differentiate between momentum and moment. And I still can't get why people here call it momentum and where this term comes from. Aug 7, 2021 at 23:21
• Oh wow interesting. I didnt know any of that Aug 8, 2021 at 1:37
• @IvanNepomnyashchikh Same confusion in Germany. It is "Impuls" in German what is "momentum" in English. Aug 8, 2021 at 8:03
• The etymology of English momentum is from “movimentum”. Not germanic side of english; is latin side of english. So it’s base was movere in Latin and it lost the “v” somewhere. But same with moment. Moment comes from movimentum also. Someone long ago thought “a moment” in time had something to do with movement somehow Aug 8, 2021 at 8:12
• Therfore the surprise is not the word ‘momentum’ from movement. The surprise is the word ‘moment’ from movement. (!) .. But the impulse force and “impulse” comes from germanic side of english. As Thomas just proved Aug 8, 2021 at 8:14