# Example: Oscillator calculations

## Understanding the situation

The simple harmonic oscillator leads itself to a number of different kinds of calculations and problems. Let's consider one as an example of how to do one. The first two parts are a standard problem in traditional physics classes. The last part gives a good example of how estimation skills play a critical role in creating first level simple models of a real physical situation.

## Presenting a sample problem

A block of mass M at rest on a horizontal frictionless table. It is attached to a rigid support by a spring of constant k. A clay pellet having mass m and velocity v strikes the block as shown in the figure and sticks to it.

1. Determine the velocity of the block immediately after the collision.
2. Determine the amplitude of the resulting simple harmonic motion.
3. In order for you to solve this problem you must make a number of simplifying assumptions, some of which are stated in the problem and some of which are not. Discuss the approximations which you had to make in order to solve the problem. (I can think of at least five.)

## Solving this problem

Although the question about the approximations is stated last, you really have to decide first how you are going to treat the problem. Unfortunately, this is often an unstated feature of many physics problems: they intend for you to solve a toy model so you can focus on some of the key mathematical features and relationships without getting messed up by what may be small (and perhaps misleading) effects. Let's therefore look at this carefully first.

iii. In the problem it explicitly mentions a couple of simplifying assumptions:

• The table is frictionless. This means we do not have to consider the force of friction acting on the block.
• The support (the vertical bar to which the right end of the spring is attached) is rigid. This means that we don't have to worry about the support bending. That would change how we measured the length of the spring. If the support is fixed we can take the 0 of position to be at the support and then the length of the spring is just given by the position of the (hook to which the spring is attached on the) block.

Well that's great, but the question suggest there are others. Here are five simplifying assumptions that are also appropriate for us to make.

• The table on which the block is sliding is horizontal. If that's the case and the table is frictionless, then the normal force from the table must cancel the object's weight while the block is anywhere on the table. We therefore don't need to consider those forces. This is probably a pretty good assumption since we can make an effort to level the table.
• The spring is an ideal Hooke's law spring (force it exerts is given by $F = -kΔL$ where the minus means that the direction of the force is to reduce the size of $ΔL$) under both extension and compression. This is suggested by the statement in the problem "a spring of constant $k$". Only an ideal Hooke's law spring can be described this way. This is probably the worst assumption we have to make. Not very many springs behave this way. Real springs tend to behave by Hooke's law when they are extended somewhat but not two much. (Having two extended springs holding the block and pulling in opposite directions is a good way to create a system that behaves like an ideal Hooke's law spring.)
• The pellet is shown as moving to the right towards the block. A real pellet would fall and follow a parabolic path. If the pellet is moving pretty fast and was launched pretty close to the block, it won't have fallen very much by the time it hits the block. So we'll just ignore gravity acting on the pellet and assume it moves in a straight line.
• Both the pellet and block are moving in air. As a result, there will be inertial drag acting on both of them. But the inertial drag is proportional to the area of the object pushing through the air and to the square of its velocity. For the block, we assume it's moving slowly (compared to its terminal velocity in air) so the drag force acting on it will be small. For the pellet, it looks like it has a small area and it won't be travelling very far. It's probably also not going anywhere near its terminal velocity in air (unless it's actually a bullet). The $v$ is shown as being its value when the pellet is pretty close to the block so we'll assume the air won't slow it down significantly before it hits the block.
• Finally, think about when the pellet hits the block. When it does, they will exert forces on each other; the pellet will slow down and the block will speed up. As the block starts to move, the spring starts to compress. The block's motion will depend on a combination of the forces from the pellet and the spring. Since we have no way to figure out the force the pellet is exerting on the block, it looks as if we are doomed! It's much too complicated. But the picture gives us a hint. The pellet looks MUCH smaller than the block. By Newton's 3rd law they exert equal (and opposite) forces on each other. But by Newton's 2nd law, the acceleration of the objects are each given by $a = F^{net}/m$. Since their forces are the same but the block's mass is MUCH bigger, the pellet's acceleration is much bigger. It stops before the block has a chance to get moving (and compress the spring and mess everything up). So it's plausible to assume that the pellet hits the block and gets absorbed by the block before the block has moved noticeably.

I'll bet you didn't expect there were that many hidden assumptions! But for a block and pellet that look like the picture and have "normal everyday" parameters -- say the block is a few inches across and has a mass of a few hundred grams, the pellet is a few grams and maybe moving a few meters per second, and the spring is a typical spring that you can stretch with your two hands -- these are reasonable assumptions. A lot of these you might have done automatically without thinking if you had a good realistic picture of what kind of system this what and what it looked like physically. If you imagined a fast pellet shot from a few inches away you wouldn't even have thought about it falling. You know from your intuition that it doesn't fall much. This is one reason we want you to learn to use your intuitions about the world to build quantitative estimations! It's not just for estimation problems -- it's hidden in almost every problem you solve.

Now let's start on the standard problems.

i. How fast is the block moving immediately after the collision? Since we have assumed that the block has all balanced forces on it when it is struck and doesn't move much while it captures the pellet, the only unbalanced force the block feels while it is struck is from the pellet. And since we are ignoring gravity on the pellet, the only unbalanced force it feels in from the block. A perfect situation for Newton's 3rd law and its consequence, momentum conservation

The initial momentum of the system is $mv$ (pellet only since block has velocity 0). After collision, let the velocity of the combined block and pellet be $V$. By momentum conservation we must have initial total momentum = final total momentum or

$$mv = (m+M)V$$

so

$$V = \frac{m}{m+M}v.$$

ii. Determine the amplitude of the resulting simple harmonic motion. Since after the collision, our assumptions say that the only unbalanced force on the block is that of the spring, it will undergo simple harmonic motion. We know that the general equation that describes this motion is (see mass on a spring)

$$x(t) = A \cos{(\omega_0 t - φ)}$$

where  $\omega_0 = \sqrt{k/m}$.  This is the general solution. It can describe any starting condition by choosing the appropriate values of $A$ and $φ$. How can we figure out what they are here?

Think about what's happening. When the pellet hits the box it gives the box a velocity $V$. So the box starts its oscillation at position $x(0)=0$ and velocity $v(0)=V$. This will be its maximum velocity and it should be positive if we take the origin at the equilibrium position of the box and the positive direction of the x axis to the right.

So we want the velocity to oscillate between $V$ (known) and $-V$, and the position to oscillate between $A$ (not yet known) and $-A$.

Since we know the starting velocity, we can differentiate our solution to get

$$\frac{d}{dt}x(t) = v(t) = -\omega_0 A \sin(\omega_0 t - φ)$$

so the maximum velocity is $ωA$. We can therefore find $A$ by writing

$$V = ω_0A$$

so

$$A = V/ω_0.$$

Since we know $ω_0$ in terms of the givens, this solves for $A$.

To get the full oscillatory solution we have to find $φ$. We want our solution for $V$ to look like $v(t) = V \cos{ (ω_0t)}$ so it will start at positive maximum velocity. This means we want to choose $φ$ so (writing $θ$ to stand for $ω_0t$ to make it look more like a math equation) we get that we have to have

$$-\sin{(θ - φ)} = \cos{(θ)}.$$

We can do this geometrically by looking at the graphs of sin and cos or by expanding out using a trig identity ($\sin{(A+B)}=\sin{A}\cos{B}+\cos{A}\sin{B}$). The graph of $-\sin{(θ)}$ and $\cos{(θ)}$ look like this:

So you have to slide the -sin curve over by 1/4 of an oscillation to the right to make it match with cos. This means we want $φ=π/2$ giving our solution as

$$x(t) = \frac{V}{ω_0} \sin{(ω_0 t)}.$$

This makes sense since now $x(t)$ starts at 0 and increases just as we need. (Or we could have skipped the math and just made this observation!)

Joe Redish 4/7/16

Article 668