Need help with Simple Harmonic Motion Problem

  • #1

Homework Statement

Part 1:
A block of mass 0.3 kg is attached to a
spring of spring constant 23 N/m on a fric-
tionless track. The block moves in simple har-
monic motion with amplitude 0.2 m. While
passing through the equilibrium point from
left to right, the block is struck by a bullet,
which stops inside the block.
The velocity of the bullet immediately be-
fore it strikes the block is 68 m/s and the mass
of the bullet is 4.67 g.
Find the speed of the block immediately
before the collision.
Answer in units of m/s.

Part 2:
If the simple harmonic motion after the collision is described by x = B sin(wt + O), what
is the new amplitude B?
Answer in units of m.

Homework Equations

Part 1:
.5(m)(v^2) + .5(k)(x^2) = .5(k)(A^2)

Part 2:
w = 2 (pie) (frequency) = [(2)(pie)]/(Period)
frequency = (2)(pie)(square root[k/mass])

The Attempt at a Solution

Part 1:
I have some ideas on how to solve the problem, but just need some guidance on if I'm doing it right and if my thinking process is right.
Since they're wanting me to find the speed of the block before the the bullet impacts it, then would I just neglect the information about the bullets?
And in the equation, how would you find x? Would you just halve the amplitude to find the x?

Part 2: The O is supposed to be a O with a vertical line in the middle of it. Not sure what that letter is called, but it looks like the greek letter, Phi. Tell me if I'm thinking this out right. I can use the frequency formula for a spring and solve for "f" and plug it back into the omega formula. Once I get the Omega, I'll plug that number into the simple harmonic motion equation. Is this right so far? But when using "m" in the frequency formula, what value of mass should I use? I'm thinking about combining the two masses, block + bullet, because the bullet is embedded into the block, creating a new system. Is this right also?
But after doing all this, how would I go about and solve for the amplitude, B. For what value should I use for "x"?
Last edited:
  • #2
Part one is REALLY simple if you just consider momentum.

Part two takes a bit more work. But basically it's at the equlib. point when velocity is max, so all the bullet has done is changed the max velocity of the block. Using the new max velocity you just need to work backwards to get the new amplitude.

Good Luck!
Last edited:
  • #3
Ok so for part 1, I used momentum.
But I need to see if this is correct before actually entering it into my homework.
I figured it to be momentum before the collision and momentum after the collision. My mass 1, was the mass of the block. Mass 2 was the mass of block + bullet, in kilogram. My velocity 1 is the unknown and velocity 2 is the velocity of the bullet. Then I solved for Velocity 1. But when I solved for the velocity, I got 69.059 m/s. This value doesn't seem right though. Shouldn't it be lower than what I actually got?

Part 2:
Tell me if this is correct so far.
I need to find the new max velocity after the bullet is embedded into the block so to do that, I used the momentum formula and solved for V2. My Mass 1 was the mass of bullet and Velocity 1 was the velocity of the bullet before impact. My Mass 2 was the mass of block + bullet and my Velocty 2 is my unknown. Basically, what I'm thinking right now is that Velocity 2 is the speed of the block with the bullet embedded in it. Is this right? After finding the new max velocity, I derived the simple harmonic equation given to me, x = B sin(wt + O), to get the velocity function, v = wBcos(wt + O). I plugged in the velocity and this is where I'm stuck. I solved for w from the equation that I listed in the "Relevant Equation," but how would I solve for Phi, the O?
Last edited:
  • #4
Yes, for the first part you calculated something wrong. Make sure everything is in kilograms
mbvb +mwoodvwood = (mb+mwood)v
And be careful not to make a sign mistake with this. Momentum is a vector.

That's the max velocity.
  • #5
I know we're trying to solve for the velocity of the wood, but I have to first find the velocity of the block + wood first right? To do this, would you set up another momentum equation. mbvb = (mb+mwood)v and solve for the velocity of the wood plus bullet?
  • #6
Wood and block are the same thing? When I used the subscript b I was meaning the bullet. Yes you do need the initial velocity of the wood, that is given by the equation I posted. And then to get the new amplitude once you have the new velocity you can use that same equation.

As for your question about the momentum, the equation I gave you is all you need
  • #7
Sorry about that: I meant bullet + wood, not block + wood. But the thing that confuses me is that we do not know the velocity of the block after the bullet lodges itself inside of it. From the question, you're only given the velocity of the bullet before impact and nothing else. How would you solve for the block+bullet velocity then? Would you assume that the velocity of the block + wood equal the velocity of the bullet since they're on a frictionless surface?
  • #8
Ok, so you only have the velocity of the bullet, but you can get the velcity of the block (before the collision) with v=wA.

Think of it as a system with two objects. The total momentum of those two objects before they collide must be the same after they collide. So the inital is just the vbmb+vwmw expression. After the collision they are both moving together with the same final velocity, so final momentum is given by (mb+mw)v. You have everything to solve for that velocity, and once you know that velocity you can find the new amplitude.
  • #9
Thanks for the help. I got the answer, but can I ask you one more question. There's a Part III also and here it is:
The collision occurred at the equilibrium position.
How long will it take for the block to reach
maximum amplitude after the collision?
Answer in units of s.
  • #10
OK, with simple harmonic motion you should know that the T, the period, is the time for one oscillation, and half the period is the time for half an oscillation, and so on. From equilibrium to the next max is what fraction of the period?

And for calculation the period there is an equation that is given on the College Board AP Equation Sheet. And notice that it is mass dependent.

Good luck!
  • #11
Ok so since the location at equilibrium is at the max, the time required to reach the next max would just be the next period. With this, I used the equation: T = (2 pie) (square root[m/k]). I got the wrong answer though, .723154006 seconds.
  • #12
You've practically got it. You're just confusing max amplitude with max velocity. Yes, it is at max velocity when it gets shot (equilibrium), but it's T/4 from the max amplitude because it's traveling in the positive x direction.
  • #13
Wait, can you explain to me what max velocity and max amplitude are? I though max amplitude was at the peak and since equilibrium also began at the peak, wouldn't the period just be 1?
  • #14
Wait tell me if this thinking process is correct:
The period from the max amplitude is T/4 because amplitude is the distance between the max/min from the middle. And since there are 4 of these distance from the middle, one from the max and middle, one from minimum and middle, another from minimum and middle, and another max from middle, then the time required to reach the next maximum amplitude would be four times? But wait. If this is the right process, then wouldn't the Period be 4T?
  • #15
I dunno, you're getting it, but why multiply by 4, that means it's going to complete for cycles.


It's staring at the equlib position, and one whole period T would be

If that makes sense; I'm trying to show you graphically. It will really help you to read this stuff out of a book though.

Anyway, that first line, in the diagram is from equlib to max amplitude, that's why you want and that's 1/4 of a cycle, and thus 1/4 of the amplitude

Suggested for: Need help with Simple Harmonic Motion Problem