Simple harmonic motion problem

In summary, the mass on a spring is oscillating and has maximum speed at 1.5m/s. At 0.60m from equilibrium, the mass has a speed of 1.875m/s.
  • #1
speedskater
9
0

Homework Statement



A mass is oscillating on a spring with simple harmonic motion. Its amplitude is 0.80m and its maximum speed is 1.5m/s, at the point of equilibrium. What is the speed of the mass at 0.60m?

Is there enough information to answer this question? The answer given is 0.75m/s, but I can't figure out how to get there.

Homework Equations



Ep = kx2/2
Ec = mv2/2
Em = Ep + Ec

The Attempt at a Solution



I've made many attempts at this, all completely wrong. I always end up trying to factorize (my teacher told us that the answer involves factorizing) but it goes absolutely haywire.

Em = Em'
Epmax = Ecmax
kxmax2/2 = mvmax2/2 (the 1/2 cancels for each formula)
vmax/xmax = √k/m (which should be constant)
(1,5m/s)2/(0.80m)2 = √k/m
1.875(not sure what unit to put here) = √k/m ∴ k/m = 3.516

kx2 + mv2 = mvmax2 (again the 1/2 cancels)
mvmax2 - mv2 = kx2
m(vmax2 - v2) = kx2
vmax2 - v2 = kx2/m
k/m = (vmax2 - v2)/x2

If the k/m stays constant, which I believe it should, it should be an easy substitution to find the v I'm looking for.

After manipulating the formula I got, I'm left with

v = √-(k/m - vmax2/x2)(x2)

After substituting, I get 0.99m/s.
 
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  • #2
Instead of all that, think in terms of fractions or ratios. At the point in question, what fraction of the total energy is Ep and what fraction is KE?
 
  • #3
Is it possible the system is hanging?I got the same answer of .99m/s using a horizontal system and energy conservation principals.
 
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  • #4
Me too.
v=VmaxSin(ArcCos(6/8))
V=1.5 X sin(41.41°)
v=0.99 m/s
http://img818.imageshack.us/img818/3757/shmw.jpg
 
Last edited by a moderator:
  • #5
I get the same answer.
 
  • #6
Doc Al said:
Instead of all that, think in terms of fractions or ratios. At the point in question, what fraction of the total energy is Ep and what fraction is KE?

Do you mean x2/xmax2? = .36/.64 = 0.5625? So 56% of the mechanical energy is potential energy? Which would mean that 44% is kinetic energy. Not sure if that's what you mean or not, or if I just did some completely useless calculations. If that's what you meant, I don't see how that will help me :/
 
  • #7
Just an observation, but [itex]1.5\sqrt{\frac{.8 - .6}{.8}} =.75[/itex], maybe that can help lead to an answer.

Like (v1/v2)2 = d1/d2, where d is the distance from the maximum amplitude, but how to prove this?
 
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  • #8
speedskater said:
Do you mean x2/xmax2? = .36/.64 = 0.5625? So 56% of the mechanical energy is potential energy? Which would mean that 44% is kinetic energy. Not sure if that's what you mean or not, or if I just did some completely useless calculations. If that's what you meant, I don't see how that will help me :/
That's exactly what I meant. It's just a quicker way of getting the (same) answer:
KEx=0.6 = .4375*KEmax
1/2mV2 = .4375(1/2mVmax2)
 
  • #9
Doc Al said:
That's exactly what I meant. It's just a quicker way of getting the (same) answer:
KEx=0.6 = .4375*KEmax
1/2mV2 = .4375(1/2mVmax2)

But if you do that (mass cancels, 1/2 cancels), you get 0.99m/s again!
 
  • #10
Another formula
ω=Vmax/R
v=ω√(A2-x2)
v=(1.5/0.8)√(0.82-0.62)
v=0.99 m/s
 
  • #11
speedskater said:
But if you do that (mass cancels, 1/2 cancels), you get 0.99m/s again!
Of course. I'd say that was the correct answer.

What book are you using? Are you sure you didn't just read the wrong answer?
 
  • #12
It's the Alberta high school physics textbook. It says 0.75m/s for sure. If all of you, who are experts in physics, agree with me and say it should be 0.99m/s, should I risk it and go with that answer? Or make up a way to get to 0.75m/s (as the rest of my class did) and just get partial points?
 
  • #13
Maybe you can show us your friend's work(rest of the class) that lead to the answer v=0.75 m/s
 
  • #14
All they did was t=d/v, using 0.80m for d. Then they manipulated the formula d=((Vi + Vf)/2)t to find Vf, using 0.60 as d. They got the answer by fluke, because velocity and acceleration are always changing in a spring.
 
  • #15
speedskater said:
All they did was t=d/v, using 0.80m for d. Then they manipulated the formula d=((Vi + Vf)/2)t to find Vf, using 0.60 as d. They got the answer by fluke, because velocity and acceleration are always changing in a spring.
Well, that 'solution' is clearly bogus. I say stick with your correct solution and trust the instructor to realize that the given answer is incorrect.
 
  • #16
It turns out that 0.99m/s is right! Thanks everyone for all your help!
 

1. What is Simple Harmonic Motion (SHM)?

Simple Harmonic Motion is a type of periodic motion in which the restoring force is directly proportional to the displacement from equilibrium position and acts in the opposite direction of the displacement. This results in a sinusoidal motion.

2. What is the equation for Simple Harmonic Motion?

The equation for Simple Harmonic Motion is x = A sin(ωt + φ), where x represents the displacement, A represents the amplitude, ω represents the angular frequency, and φ represents the phase angle.

3. How is Simple Harmonic Motion different from other types of motion?

Simple Harmonic Motion is different from other types of motion because it follows a specific pattern of oscillation that is dependent on the restoring force and displacement from equilibrium. Other types of motion, such as linear or circular motion, do not exhibit this pattern.

4. What is the relationship between Simple Harmonic Motion and energy?

In Simple Harmonic Motion, there is a constant exchange of potential and kinetic energy. As the object moves away from equilibrium, it gains potential energy, which is then converted to kinetic energy as it moves towards equilibrium. This process continues as the object oscillates back and forth.

5. How is Simple Harmonic Motion applied in real-life situations?

Simple Harmonic Motion can be observed in many real-life situations, such as the motion of a pendulum, a mass-spring system, or a vibrating guitar string. It is also used in various engineering and scientific applications, such as in the design of shock absorbers, tuning forks, and musical instruments.

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