What Is the Speed of an Oscillating Particle at Equilibrium?

[goonking]

Homework Statement

Homework Equations

The Attempt at a Solution

the particle is oscillating

so we know at 1.5 seconds, it is at the equilibrium point , where speed is highest. (the speeds are 0 at the highest and lowest points of the graph.)

we see that it goes from -6m to 6m, which is 12 meters in 1 second.

so is the answer 12 meters per second? since that is the highest it can go?

 

[AlephNumbers]

Are you familiar with the equations that govern simple harmonic motion?

 

[goonking]

Are you familiar with the equations that govern simple harmonic motion?

no sir, i will google it right now.

 

[SammyS]

Homework Statement

Homework Equations

The Attempt at a Solution

the particle is oscillating

so we know at 1.5 seconds, it is at the equilibrium point , where speed is highest. (the speeds are 0 at the highest and lowest points of the graph.)

we see that it goes from -6m to 6m, which is 12 meters in 1 second.

so is the answer 12 meters per second? since that is the highest it can go?

That's just the average speed from t = 1 second to t = 2 seconds .

 

[AlephNumbers]

I would check out hyperphysics. http://hyperphysics.phy-astr.gsu.edu/hbase/shm.html

 

[goonking]

I would check out hyperphysics. http://hyperphysics.phy-astr.gsu.edu/hbase/shm.html

yes, I am currently reading that.

 

[goonking]

I would check out hyperphysics. http://hyperphysics.phy-astr.gsu.edu/hbase/shm.html

so we need to use v = ω A cos ωt ?

 

[AlephNumbers]

so we need to use v = ω A cos ωt ?

Yes.

 

[goonking]

Yes.

how do we find angular velocity? Is the particle moving around in a circle? I thought it was more like a spring, going back and forth.

 

[AlephNumbers]

I thought it was more like a spring, going back and forth.

It is.

Is the particle moving around in a circle?

No, but its horizontal displacement is the same as if it were moving in uniform circular motion. I would look at a textbook or online resource for a more complete explanation.

how do we find angular velocity?

The distance between two crests (high points) on the graph you provided is the amount of time it takes the mass to make a complete "revolution" (2π radians). Angular velocity is measured in radians per second... do you see where I am going with this?

 

[goonking]

yes, so it takes 2 seconds to do 1 rev.

so 2π radians in 2 secs , 1 radian per sec.

ω = 1 radian per sec.

A = 3 meters

and cos ω t = cos ((1r/s) (2 second)) = .99939
v = ω A cos ωt = 1 x 3 x .999 = 3 m/s

correct?

 

[AlephNumbers]

so 2π radians in 2 secs , 1 radian per sec.

On a bit of a physics binge, eh? Don't worry, PF has you covered.

 

[BvU]

Draw a straight line with slope 3 m/s through the point (1.5s, 0 m) to check...

 

[goonking]

(2π)/2 = π radians per second.

On a bit of a physics binge, eh? Don't worry, PF has you covered.

yes, silly me always doing homework when everyone is asleep.

 

[BvU]

Well done !

 

[goonking]

(2π)/2 = π radians per second.

whoops, yes.

and cos ω t = cos ((π r/s) (2 second)) = .9984

answer still came out to be 3 m/s.

 

[goonking]

Draw a straight line with slope 3 m/s through the point (1.5s, 0 m) to check...

like this?

I have no idea how to confirm if that is suppose to correct or not.

 

[AlephNumbers]

No, I think BvU meant more like this

 

[goonking]

No, I think BvU meant more like thisView attachment 82264

i think he was trying to tell me my answer was wrong.

by the looks of it, my speed should be -3m/s instead of 3 m/s.

correct?

 

[AlephNumbers]

i think he was trying to tell me my answer was wrong.

by the looks of it, my speed should be -3m/s instead of 3 m/s.

correct?

You just drew a tangent line at t = 2.5 instead of t = 1.5. At t = 2.5 the mass has a negative velocity. I believe BvU was just pointing out that you could draw a line tangent to the graph in order to check your answer.

 

[goonking]

No, you solved the problem. You just drew a tangent line at t = 2.5 instead of t = 1.5. At t = 2.5 the mass has a negative velocity. I believe BvU was just pointing out that you could draw a line tangent to the graph in order to check your answer.

oh wow, i meant to draw at 1.5 sec, not 2.5 sec, truly sorry!

 

[goonking]

oh wow, i meant to draw at 1.5 sec, not 2.5 sec, truly sorry!

the answer should be 3 m/s then , right?

 

[AlephNumbers]

yes, so it takes 2 seconds to do 1 rev.

so 2π radians in 2 secs , 1 radian per sec.

ω = 1 radian per sec.

A = 3 meters

and cos ω t = cos ((1r/s) (2 second)) = .99939
v = ω A cos ωt = 1 x 3 x .999 = 3 m/s

correct?

Your value of A is not correct. It should be the maximum displacement. Also, make sure to use the correct value for ω.

 

[goonking]

Your value of A is not correct. It should be the maximum displacement. Also, make sure to use the correct value for ω.

sorry, let me fix that.

A = 6 meters.
ω = π r/s

correct?

 

[AlephNumbers]

sorry, let me fix that.

A = 6 meters.
ω = π r/s

correct?

Looks good!

 

[goonking]

Looks good!

getting an answer to be 18.7 m/s.

if i want to check that, I would need to graph a line with slope 18.7?!

 

[AlephNumbers]

Yes. If you draw a line tangent to the graph at t = 1.5 you should get a line that has a slope of about 18.7

 

[goonking]

Yes. If you draw a line tangent to the graph at t = 1.5 you should get a line that has a slope of about 18.7

awesome, thanks!

 

[BvU]

The line you drew is going through (2.5 s, 0 m) and has a slope 18 m/s.
Two out of three ain't bad, but here only one out of three values is as intended.

Coincidentally the 18 m/s is close to the correct answer of $A\omega$ where you found the wrong value for two out of two. That's not good at all. Fortunately, you already fixed the $\omega = {2\pi\over 2} \ne 1$.

Fixing a factor of $\pi$ corrects one error in your answer. Guess what's wrong with the other...

Perhaps now you can draw a line that is going through (2.5 s, 0m) and has a slope 6$\pi$ m/s to check that that is indeed the correct answer .

 

[goonking]

The line you drew is going through (2.5 s, 0 m) and has a slope 18 m/s.
Two out of three ain't bad, but here only one out of three values is as intended.

Coincidentally the 18 m/s is close to the correct answer of $A\omega$ where you found the wrong value for two out of two. That's not good at all. Fortunately, you already fixed the $\omega = {2\pi\over 2} \ne 1$.

Fixing a factor of $\pi$ corrects one error in your answer. Guess what's wrong with the other...

Perhaps now you can draw a line that is going through (2.5 s, 0m) and has a slope 6$\pi$ m/s to check that that is indeed the correct answer .

why does it have to be 2.5 s? don't you mean 1.5 s?

 

[BvU]

PS personally I would never trade in the perfectly correct value of $6\pi$ m/s for an approximate answer (18.7 m/s w is a wrong rounding off for 18.8495559215... ) unless I was really forced to do so. After all, in subsequent calculations factors $\pi$ may well cancel out.

 

[goonking]

PS personally I would never trade in the perfectly correct value of $6\pi$ m/s for an approximate answer (18.7 m/s w is a wrong rounding off for 18.8495559215... ) unless I was really forced to do so. After all, in subsequent calculations factors $\pi$ may well cancel out.

very true, ill keep that in mind

 

[BvU]

Draw a straight line with slope 3 m/s through the point (1.5s, 0 m) to check...

Where did the 2.5 s come from ? Not from me.

 

[goonking]

Where did the 2.5 s come from ? Not from me.

oh, i thought i needed to check by plotting the line at 2.5 secs with slope 6 pi.

but yes, before the 2.5 was a mistake i made :(

 

[BvU]

I'm lagging with my slow typing.

Acquiring some "dexterity" with sines and cosines is a good idea.
This exercise helps, but the $A$ and $\omega$ obfuscate things a bit (intentionally, from the point of viewof the exercise composer). For you, practicing with $x = \sin (\omega t)$ is more helpful (i.e. A = 1 and $\omega = 1$).

Draw a graph of that and a unit circle on the same scale to the left and tadaa: values of x and speed at the angles $0, {\pi\over 6}, {\pi\over 4}, {\pi\over 3}, {\pi\over 2}, {2\pi\over 3}, {5\pi\over 6}, {\pi}$ and each of these + ${\pi}$ become clear.

See how they all hang together, and also hang together with ${d^2x\over dt^2} = -x$.

Once you have that internalized, dealing with $A\ne 0$ and $\omega \ne 0$ is a piece of cake and your efficiency in excercises will improve; also: you don't have to remember all that much.

 

[goonking]

I'm lagging with my slow typing.

Acquiring some "dexterity" with sines and cosines is a good idea.
This exercise helps, but the $A$ and $\omega$ obfuscate things a bit (intentionally, from the point of viewof the exercise composer). For you, practicing with $x = \sin (\omega t)$ is more helpful (i.e. A = 1 and $\omega = 1$).

Draw a graph of that and a unit circle on the same scale to the left and tadaa: values of x and speed at the angles $0, {\pi\over 6}, {\pi\over 4}, {\pi\over 3}, {\pi\over 2}, {2\pi\over 3}, {5\pi\over 6}, {\pi}$ and each of these + ${\pi}$ become clear.

See how they all hang together, and also hang together with ${d^2x\over dt^2} = -x$.

Once you have that internalized, dealing with $A\ne 0$ and $\omega \ne 0$ is a piece of cake and your efficiency in excercises will improve; also: you don't have to remember all that much.

thanks for the advice :)

 

[BvU]

Welcome. Good luck with your physics binge.

More advice: get some rest when wearing out !

 

[goonking]

Welcome. Good luck with your physics binge.

More advice: get some rest when wearing out !

hehe, no rest for me.