Time Between max speed and max acceleration on a spring

  • Thread starter Hypnos_16
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  • #1
Hypnos_16
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Homework Statement



An object attached to a horizontal spring is oscillating back and forth along a frictionless surface. The maximum speed of the object is 1.16 m/s, and its maximum acceleration is 6.52 m/s2. How much time elapses between an instant when the object's speed is at a maximum and the next instant when its acceleration is at a maximum?

Vmax = 1.16m/s
Amax = 6.52m/s^2


Homework Equations



I'm not sure of any equations that would compare these two variables, or that would cancel one or another out.


The Attempt at a Solution



I don't have an attempt at this one. Sorry.
 

Answers and Replies

  • #2
gneill
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What sort of motion does a mass-spring system undergo?
 
  • #3
tiny-tim
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Hi Hypnos_16! :smile:

(have an omega: ω and try using the X2 icon just above the Reply box :wink:)

Hint: what is the general equation for such a motion? :wink:
 
  • #4
Hypnos_16
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Well for max Speed it's v = Aω and for Acceleration it's a = Aω2 or at least, those are the two we've been taught
 
  • #5
tiny-tim
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Exactly! :smile:

soooo … ? :wink:
 
  • #6
gneill
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Well for max Speed it's v = Aω and for Acceleration it's a = Aω2 or at least, those are the two we've been taught

Yes, but where did they come from? What's the equation of motion?
 
  • #7
Hypnos_16
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Speed is v = -Awsin(wt)
Acceleration is a = -Aw2cos(wt)
Okay, but how does this help? I don't see how anything would cancel, or how this benefits me.
 
  • #8
berkeman
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Speed is v = -Awsin(wt)
Acceleration is a = -Aw2cos(wt)
Okay, but how does this help? I don't see how anything would cancel, or how this benefits me.

Drawing a graph should help you figure it out.

Draw a single vertical axis, and three horizontal axes through it. On the top horizontal axis, graph position as a function of time. On the 2nd horizontal axis, graph the equation you show for v(t). And on the bottom axis, graph a(t).

Now do you see where the max value(s) of each is, and why it is there?
 
  • #9
tiny-tim
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  • #10
Hypnos_16
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soooo … ??

then i assume things will start to cancel
a = aω2
6.52m/s2 = aω2
so if i square root both sides i have
2.55m/s = aω


and for speed
v = aω
so 1.16m/s also = aω
so.... have i gone completely off track here?
 
  • #11
tiny-tim
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6.52m/s2 = aω2
so if i square root both sides i have
2.55m/s = aω

nooooo :redface: :cry:
 
  • #12
Hypnos_16
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Then help me out! i'm not getting it.
 
  • #13
tiny-tim
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You have Aω, and you have Aω2,

so ω = … ?​
 
  • #14
gneill
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then i assume things will start to cancel
a = aω2
6.52m/s2 = aω2
so if i square root both sides i have
2.55m/s = aω


and for speed
v = aω
so 1.16m/s also = aω
so.... have i gone completely off track here?

Gaaaa! Why are you using "a" for two different variables? It's a and A!

Write your two equations. You're given a and v. Solve for A and ω. Or just ω, since that's really all you'll need to move forward.
 
  • #15
Hypnos_16
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sooooo would it be Aω2 / Aω
leaving just a ω left over?
 
  • #16
tiny-tim
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yes!! :biggrin:

is everything ok now?
 
  • #17
gneill
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sooooo would it be Aω2 / Aω
leaving just a ω left over?

Sure. Now, what are you going to do with that? Can you put a value to it?
 
  • #18
Hypnos_16
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Okay so i have ω now, and when i sub that into the equation of v = -Aωsin(ωt), what do i put as my v? I have A and ω and everything, but if i use the max speed for v than it's going to be odd.
 
  • #19
gneill
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Okay so i have ω now, and when i sub that into the equation of v = -Aωsin(ωt), what do i put as my v? I have A and ω and everything, but if i use the max speed for v than it's going to be odd.

You first have to decide what it is that you want to find next. Go back tot he problem statement and see what it's looking for.
 
  • #20
tiny-tim
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… but if i use the max speed for v than it's going to be odd.

You don't need the max speed for v, you only need the times.
 
  • #21
Hypnos_16
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I got the max speed took 0.279s to reach, and i'm trying to do the acceleration, but when i try using a = -Aω2cos(ωt) i keep getting 1 = cos(5.62t) and the cos-1 of 1 is 0. What have i messed up this time?
 
  • #22
gneill
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I got the max speed took 0.279s to reach, and i'm trying to do the acceleration, but when i try using a = -Aω2cos(ωt) i keep getting 1 = cos(5.62t) and the cos-1 of 1 is 0. What have i messed up this time?

You need to take a small step back from working the math and consider the physical situation and how it relates to the formulas you're using.

The expressions for the velocity and acceleration are sine and cosine curves, right? At what angles do these functions hit their maximum magnitudes (that is, +/- extremes)?

What is the angular offset between when one is maximized and the other is maximized?
 
  • #23
tiny-tim
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Hi Hypnos_16! :smile:

(just got up :zzz: …)

You're over-thinking this!

As gneill :smile: says …
You need to take a small step back …

Draw the curve, then pretend you're explaining the question to a six-year-old kid (no maths)! :wink:
 
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