# Time Between max speed and max acceleration on a spring

## 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.

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gneill
Mentor
What sort of motion does a mass-spring system undergo?

tiny-tim
Homework Helper
Hi Hypnos_16!

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

Hint: what is the general equation for such a motion?

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

tiny-tim
Homework Helper
Exactly!

soooo … ?

gneill
Mentor
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?

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.

berkeman
Mentor
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.

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?

tiny-tim
Homework Helper
Vmax = 1.16m/s
Amax = 6.52m/s^2
Well for max Speed it's v = Aω and for Acceleration it's a = Aω2
soooo … ??

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?

tiny-tim
Homework Helper
6.52m/s2 = aω2
so if i square root both sides i have
2.55m/s = aω
nooooo

Then help me out! i'm not getting it.

tiny-tim
Homework Helper
You have Aω, and you have Aω2,

so ω = … ?​

gneill
Mentor
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.

sooooo would it be Aω2 / Aω
leaving just a ω left over?

tiny-tim
Homework Helper
yes!!

is everything ok now?

gneill
Mentor
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?

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.

gneill
Mentor
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.

tiny-tim
Homework Helper
… 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.

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?

gneill
Mentor
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?

tiny-tim
Homework Helper
Hi Hypnos_16!

(just got up :zzz: …)

You're over-thinking this!

As gneill 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)!

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