Simple harmonic motion: time for maximum speed

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Homework Statement

A particle moving along the x-axis in simple harmonic motion starts from its equilibrium position, the origin, at t = 0 and moves to the right. The amplitude of its motion is 3.20 cm, and the frequency is 1.40 Hz.

Determine the maximum speed of the particle.

Determine the earliest time (t > 0) at which the particle has this speed

Homework Equations

V max = ωA

f = 1/T

T = 1/f

ω = 2∏/T

x(t) = Acos(ωt + ∅)

max. acceleration = ω²A

The Attempt at a Solution

I only have a problem finding the time at which the particle reaches the max speed, don't know where to start with that at all, but:

f = 1/T

T = 1/1.4

T = 0.714 s

ω = 2∏/ T

ω = 8.7999 rad / s

Vmax = ωA

Vmax = 0.28 m/s

Vmax = 28.1 cm/s

Where do I start with finding the time for the maximum speed?

Edit: Okay someone told me I should get the equation as a function of time for it:

which is X(t) = Acos(ωt + ∅ ) ??

so i got x(t) = 0.032cos(8.8t + ∅ )

derivative is, for v,

v = -wAsin(wt + ∅)

 

[ehild]

To make your X(t) function complete, use the information that the motion starts from equilibrium, that is, at X(t=0) = 0.
At what position is the speed maximum?

ehild

 

[Color_of_Cyan]

To make your X(t) function complete, use the information that the motion starts from equilibrium, that is, at X(t=0) = 0.
At what position is the speed maximum?

ehild

0, 2∏ ?

I honestly do not know what you are trying to imply.

But the position where the speed is maximum is I guess at the beginning of the motion, 0, 2∏ and then over and over again.

x(t) = Acos(ωt + ∅)

x(t) = 0.032cos(8.8t + ∅ )

0 = (0.032)cos[(8.8)(0) + ∅) ?
Vmax = -wAsin(wt + ∅)

0.281 = (-0.282)sin(8.8t + ∅)

Solve for t ?

0.281/ (-0.282) = sin(8.8t + ∅)

-0.996 = sin(8.8t + ∅)

sin^-1(-0.996) = (8.8t + ∅)

-1.486 = 8.8t + ∅

I think ∅ is 2∏

-1.486 = 8.8t + 2∏

t = -0.88 but this can't be right at all please help, thank you

by the way i have to do the same thing for acceleration

 

[ehild]

I honestly do not know what you are trying to imply.

But the position where the speed is maximum is I guess at the beginning of the motion, 0, 2∏ and then over and over again.

Yes, the speed is maximum at t=0. But that happens twice during a cycle. Do not forget that speed is magnitude of velocity. And you need to give the earliest time after the start when the maximum speed is reached. 2pi is not time.

x(t) = Acos(ωt + ∅)

x(t) = 0.032cos(8.8t + ∅ )

0 = (0.032)cos[(8.8)(0) + ∅) ?

What is ∅ then?

ehild

 

[Color_of_Cyan]

Yes, the speed is maximum at t=0. But that happens twice during a cycle. Do not forget that speed is magnitude of velocity. And you need to give the earliest time after the start when the maximum speed is reached. 2pi is not time.
What is ∅ then?

ehild

I meant the position not time, it seems easier to think of it that way... continuing from what you said:

0 = (0.032)cos[(8.8)(0) + ∅)

so I am solving for ∅? and cos(whatever) should equal 0 ?

in that case:

cos^-1(0) = (∏/2) ?

so ∅ = ∏/2 ?still do not really know what i am supposed to do or what you are trying to imply, but thank you, so far

 

[ehild]

What should be the value of ∅ so as x(0)=0? It is not pi.

The speed is maximum at position x=0, but x can not be 2pi.
What is the time when the particle returns next time to x=0? ehild

 

[Color_of_Cyan]

What should be the value of ∅ so as x(0)=0? It is not pi.

The speed is maximum at position x=0, but x can not be 2pi.
What is the time when the particle returns next time to x=0?


ehild

ugh

i said it was ∏/2 not ∏ or am i missing something ?

cos 90 deg = 0

90 deg -> ∏ / 2 rad

the time when the particle returns to x = 0 is the period... T = 0.714 s

but it is not what i am looking for... and i don't really see how it helps that much

what am i doing wrong??

 

[ehild]

ugh

i said it was ∏/2 not ∏ or am i missing something ?

cos 90 deg = 0

90 deg -> ∏ / 2 rad

Sorry, I misread it. Well, it is pi/2 or -pi/2 if x(0)=0. But it is said that the particle moves to the right. If moving to the right means positive velocity ∅=-pi/2.

Or it would be simpler to say that x(t) = 0.032sin(ωt).

the time when the particle returns to x = 0 is the period... T = 0.714 s

No. How many times is x=0 during a period?

but it is not what i am looking for... and i don't really see how it helps that much

what am i doing wrong??

What is you are looking for? The question is:

Determine the earliest time (t > 0) at which the particle has this speed

The particle has maximum speed at x=0. When is the particle next time at x=0?


ehild

 

[gneill]

Here's a diagram that may help to think about the problem:

 

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I finally got it right... t = 0.357s

It is just the period time in half! I might still be missing a point though;

It seems though that the equation didn't help much at all.

I had NO IDEA why you guys were trying to set the position to x=0 in the equation for it, you just could have said sooner

t = 0.357 when x = 0 because it is half of the period when x = 0 AGAIN. Thanks.

and from the graph I would say the max acc begins at the amplitude / height of it.

So instead of x = 0 it would be x = ∏/2 or x = 3∏/2 when trying to find the time to reach position of max acceleration right?

 

[ehild]

I finally got it right... t = 0.357s

It is just the period time in half! I might still be missing a point though;

It seems though that the equation didn't help much at all.

The equation should have been x(t)=0.032 sin(2.8π t).

v(t)=0.032(2.8π) cos(2.8π t).

The speed (magnitude of v) is highest when the value of the cosine is ±1, that is, when the phase is 2.8πt = 0, π, 2π, ...For the soonest after zero t=π/(2.8π)=0.357

and from the graph I would say the max acc begins at the amplitude / height of it.

So instead of x = 0 it would be x = ∏/2 or x = 3∏/2 when trying to find the time to reach position of max acceleration right?

∏/2 or 3∏/2 is not X, but the phase when the acceleration is of maximum magnitude. X= ±0.032 m there.

ehild

 

[Color_of_Cyan]

The equation should have been x(t)=0.032 sin(2.8π t).

umm... why?

how??

The equation i had in my first post was wrong though...

i have now x(t) = 3.2 cos(8.8t - pi/2)
i thought it was always x(t) = Acos(ωt + ∅) ?? seems like i just got lucky by dividing the period time in half then... ugh

∏/2 or 3∏/2 is not X, but the phase when the acceleration is of maximum magnitude. X= ±0.032 m there.

yeah thanks I forgot to see that.thanks a lot, you are good and smart

 

[ehild]

umm... why?

how??


i thought it was always x(t) = Acos(ωt + ∅) ??

It can be Asin(wt+ψ ) as well. Both are SHM, and choosing appropriate phase constant, they are the same. Asin(ωt+π/2 )=Acos(ωt), Acos(ωt-π/2)=Asin(ωt).

seems like i just got lucky by dividing the period time in half then... ugh

That dividing by half was OK, as the motion started with highest speed and there are two times when the speed is maximum during a cycle.

Next time draw a diagram before you do any calculations. Something like the one gneill posted. Or make it plot out by wolframalpha.

ehild