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Find the amplitude of the resulting simple harmonic motions

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Homework Statement


An 8.0 lb block is suspended from a spring with a force constant of 3.0 lb/ in. A bullet weighing 0.10 lb is fired into the block from below with a speed of 500 ft/sec and comes to rest in the block.
a) Find the amplitude of the resulting simple harmonic motions
b) What fraction of the original kinetic energy of the bullet is stored in the harmonic oscillator?

Homework Equations


F = -kx
m1v1 = m2v2
v = xm * ω
ω = √(k/m)
U = 0.5 kx2
K = 0.5 mv2
g = 32.2 ft/s2

The Attempt at a Solution


For a)

m1v1 = m2v2

v2 = m1v1/m2
v2 = [3.1 x 10-3 slugs * 500 ft/sec] / [0.248 slugs]
= ~6.17 ft/sec

So:
v = xm * ω
= xm * √(k/m)

xm = v * √(m/k)
= 6.2 ft/sec * √[0.25 slugs / (36.0 lbs/ft)] *I converted from 3.0 lb/in here
= 0.52 ft

...I solved that correctly, right?

and for b)
Ui = 0
Ki = 0.5m1*v12
= 0.5 * (3.1x10-3 slugs) * (500 ft/s)2
= 388 ft-lb

Kf = 0
Uf = 0.5*k*x2
= 0.5 * (36.0 lb/ft) * (0.52 ft)2
= 4.87 ft-lb

...Now, I imagine that quite a lot of energy would be "lost" as heat due to friction (and other, more complicated forces that this problem doesn't quite take into account) but this seems like a rather large fraction lost. Do I have an arithmetic error? Or did I just use the wrong formula?
 

Answers and Replies

  • #2
collinsmark
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Homework Statement


An 8.0 lb block is suspended from a spring with a force constant of 3.0 lb/ in. A bullet weighing 0.10 lb is fired into the block from below with a speed of 500 ft/sec and comes to rest in the block.
a) Find the amplitude of the resulting simple harmonic motions
b) What fraction of the original kinetic energy of the bullet is stored in the harmonic oscillator?

Homework Equations


F = -kx
m1v1 = m2v2
v = xm * ω
ω = √(k/m)
U = 0.5 kx2
K = 0.5 mv2
g = 32.2 ft/s2

The Attempt at a Solution


For a)

m1v1 = m2v2

v2 = m1v1/m2
v2 = [3.1 x 10-3 slugs * 500 ft/sec] / [0.248 slugs]
= ~6.17 ft/sec
Is m2 the mass of the block or the mass of the block+bullet combination?
 
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  • #3
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Is m2 the mass of the block or the mass of the block+bullet combination?
Mass of the bullet+block combo.
 
  • #4
collinsmark
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Mass of the bullet+block combo.
That's good.

The reason I ask is because in your first calculation, you were dividing by 0.248 slugs. That's the mass of the block. But I think it should be the mass of the bullet+block combination.

It doesn't make a huge difference, but I thought I'd point that out.

I'll comment on the rest in a following post.
 
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  • #5
collinsmark
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So:
v = xm * ω
= xm * √(k/m)

xm = v * √(m/k)
= 6.2 ft/sec * √[0.25 slugs / (36.0 lbs/ft)] *I converted from 3.0 lb/in here
= 0.52 ft

...I solved that correctly, right?

and for b)
Ui = 0
Ki = 0.5m1*v12
= 0.5 * (3.1x10-3 slugs) * (500 ft/s)2
= 388 ft-lb

Kf = 0
Uf = 0.5*k*x2
= 0.5 * (36.0 lb/ft) * (0.52 ft)2
= 4.87 ft-lb

...Now, I imagine that quite a lot of energy would be "lost" as heat due to friction (and other, more complicated forces that this problem doesn't quite take into account) but this seems like a rather large fraction lost. Do I have an arithmetic error? Or did I just use the wrong formula?
The energy lost is due to the inelastic collision between the bullet and the block.

Anyway, you have a perfectly valid approach on obtaining the final values that you did. (Assuming you fix the part mentioned in my last post about forgetting to include the mass of the bullet.)

But still, you lost some precision and accuracy along the way, one way or another, probably due to rounding errors (above and beyond the error introduced by forgetting to include the mass of the bullet). I suggest keeping more significant digits in your intermediate steps, and rounding off the proper number of digits (that your instructor asks for) as a final step.

But yes, the general methods in which you used to get those values look valid to me. So good job there. :smile:
 
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  • #6
gneill
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Mass of the bullet+block combo.
Is the equilibrium location of the bullet+block combo the same as that of the block alone? Does that tell you anything about the initial conditions of the oscillator immediately after the bullet strikes? Does your formula ##v = x_m ω## apply when the mass is not just passing through equilibrium (where the velocity should be maximum)?
 
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  • #7
collinsmark
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Is the equilibrium location of the bullet+block combo the same as that of the block alone? Does that tell you anything about the initial conditions of the oscillator immediately after the bullet strikes? Does your formula ##v = x_m ω## apply when the mass is not just passing through equilibrium (where the velocity should be maximum)?
Ooh, that's a good point. I forgot about that little detail myself. o:)
 
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  • #8
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The energy lost is due to the inelastic collision between the bullet and the block.

Anyway, you have a perfectly valid approach on obtaining the final values that you did. (Assuming you fix the part mentioned in my last post about forgetting to include the mass of the bullet.)

But still, you lost some precision and accuracy along the way, one way or another, probably due to rounding errors (above and beyond the error introduced by forgetting to include the mass of the bullet). I suggest keeping more significant digits in your intermediate steps, and rounding off the proper number of digits (that your instructor asks for) as a final step.

But yes, the general methods in which you used to get those values look valid to me. So good job there. :smile:
Yeah, when I did the problems out, I kept my significant digits (I screwed up the first few times I worked with lbs, and so when I do my work out, I usually keep it as lbs/32.2 ft/s2. I just wanted a number here, which ended up losing a lot of precision.

I'll keep that in mind in later problems. :) Thanks
 
  • #9
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Is the equilibrium location of the bullet+block combo the same as that of the block alone?
No. I think my friend was trying to explain to me that the reason I could just use Potential Energy in the equation was because the block/bullet combo is at it's maximum compression where I'm evaluating it (so, no Kinetic energy).

Does that tell you anything about the initial conditions of the oscillator immediately after the bullet strikes?
No?

Does your formula ##v = x_m ω## apply when the mass is not just passing through equilibrium (where the velocity should be maximum)?
I honestly don't know.
 
  • #10
collinsmark
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No. I think my friend was trying to explain to me that the reason I could just use Potential Energy in the equation was because the block/bullet combo is at it's maximum compression where I'm evaluating it (so, no Kinetic energy).
@gneill does have a point. Due to the slight increase in weight, the new equilibrium position of the block+bullet combination is slightly lower than the position where the block started at rest before the collision. The combination's maximum velocity will be at this new equilibrium position. Immediately after the collision, there is new kinetic energy of the mass_bullet combination (obviously), but don't forget there's also the new gravitational potential energy of the bullet (potential energy that was not present when the block was sitting there at rest, alone).

In other words, the initial velocity of the block+bullet combination, immediately after the collision, is not quite the combination's maximum velocity.

Recall that this maximum velocity was used, in part, to calculate the amplitude of the oscillation. So if we need to alter the max velocity compared the original calculations, the oscillation amplitude needs to change a little too.

----

On the other hand, this increase due to the gravitational potential energy of the bullet can be neglected for part b), since part b) specifically asks about the bullet's kinetic energy (implying the bullet's gravitational potential energy is not part of what is being asked about for that part of the problem). [Edit: at least that's the way I interpret part b). Although the more I think about it, the problem statement might be a little ambiguous regarding this.]
 
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