# Displacement of a particle

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• Jun 9th 2012, 03:17 AM
srh
Displacement of a particle
Ok I'm guessing this is simple harmonic motion?
I don't understand how these values relate to the displacement at all, can anyone help?

The displacement, d, in cm, of a particle is given by
d = 3 sin(40 π t) – 4 cos( 40 π t ) t ≥ 0
· Find expressions for the velocity and acceleration of the particle
· Calculate the velocity and acceleration when t = 2
· Determine when the particle is first at rest
· Calculate the displacement and acceleration when the particle is at rest

(π = pi)

Thanks!
• Jun 9th 2012, 03:26 AM
Goku
Re: Displacement of a particle
Find the derivative to get the velocity, and the second derivative to find the acceleration with respect to time.
• Jun 9th 2012, 04:24 AM
srh
Re: Displacement of a particle
Are you sure?
• Jun 9th 2012, 04:28 AM
Plato
Re: Displacement of a particle
Quote:

Originally Posted by srh
Are you sure?

Yes.
$v=d'~\&~a=v'=d''$
• Jun 9th 2012, 04:33 AM
Goku
Re: Displacement of a particle
• Jun 9th 2012, 04:55 AM
srh
Re: Displacement of a particle
Then when is the particle at rest?
• Jun 9th 2012, 05:17 AM
Goku
Re: Displacement of a particle
$d = 3 sin(40 nt) - 4 cos( 40 nt )$

$d^{'} = 120ncos(40nt) + 160nsin(40nt)$

Particle at rest when velocity =0.

$0 = 120ncos(40nt) + 160nsin(40nt)$

$-120ncos(40nt) = 160nsin(40nt)$

$-\frac{4}{3}tan(40nt) = 1$

$tan(40nt) = -\frac{3}{4}$

and then solve for t....
• Jun 9th 2012, 07:48 AM
srh
Re: Displacement of a particle
Thanks for your time but that doesn't help me in the slightest I'm afraid.

Can this be explained in a more simplistic way?
• Jun 9th 2012, 07:51 AM
srh
Re: Displacement of a particle
Also the π symbol is supposed to be the pi symbol
• Jun 9th 2012, 08:06 AM
Plato
Re: Displacement of a particle
Quote:

Originally Posted by srh
Also the π symbol is supposed to be the pi symbol

Use the reply with quote tab. On the tool bar you see $~~\Sigma~~$ that is the LaTeX rap.
[TEX]\pi [/TEX] gives $\pi$.
• Jun 9th 2012, 08:06 AM
Goku
Re: Displacement of a particle
Quote:

The displacement, d, in cm, of a particle is given by
d = 3 sin(40 π t) – 4 cos( 40 π t ) t ≥ 0
· Find expressions for the velocity and acceleration of the particle
So we know how to get velocity and acceleration i.e.

$velocity = 120n cos(40nt) + 160n sin(40nt)$ --> which is the derivative of displacement
$acceleration = -4800 n^2 sin(40nt) + 6400n^2 cos(40nt)$ --> which is the derivative of the velocity.

Quote:

Calculate the velocity and acceleration when t = 2
now we substitute t=2 into the above:

$velocity = 120(\pi) cos(40\pi(2)) + 160\pi sin(40\pi(2)) = 120\pi$
$acceleration = -4800 \pi^2 sin(40\pi(2)) + 6400\pi^2 cos(40\pi(2)) = 6400\pi^2$

Say the part you don't understand (Hi)
• Jun 9th 2012, 08:47 AM
srh
Re: Displacement of a particle
Any of it really :S
• Jun 9th 2012, 09:28 AM
HallsofIvy
Re: Displacement of a particle
Quote:

Originally Posted by srh
Thanks for your time but that doesn't help me in the slightest I'm afraid.

Can this be explained in a more simplistic way?

For this problem to make any sense to you at all, you have to know some Calculus. Are you taking a Calculus course? If not, what course is this for?
• Jun 9th 2012, 09:38 AM
srh
Re: Displacement of a particle
I'm studying mechanical engineering, I've got only a basic grasp of calculus from what I've been able to teach myself online.
• Jun 9th 2012, 10:24 AM
Goku
Re: Displacement of a particle
These are equations that describe the motion.

Let's take a simple example

displacement = t for $t\geq0$

For the 1st second: displacement = 1 cm
For the 2nd second: displacement = 2 cm and so on....

One can easily that the velocity is 1cm per second.
So how do we do this mathematically,
we use the formula $velocity = \frac{change\hspace{2mm} in\hspace{2mm} displacement}{change\hspace{2mm} in \hspace{2mm}time}$ = derivative of displacement
from now on I will use s to refer to displacement...

velocity = $\frac{ds}{dt}$(t)= 1 cm per second.

Now to calculate the acceleration:
You the same as above but now you take the derivative of the velocity, so we get

acceleration = $\frac{dv}{dt}$ (1)= 0.

You can verify this by seeing that the velocity is constant.

Second example

$displacement = t^2$ for $t\geq0$

For the 1st second: displacement = 1 cm
For the 2nd second: displacement = 4 cm and so on....

we use the formula $velocity = \frac{change \hspace{2mm}in\hspace{2mm} displacement}{change \hspace{2mm}in \hspace{2mm}time}$ = derivative of displacement
from now on I will use s to refer to displacement...

velocity = $\frac{ds}{dt}$ $(t^2)$= 2t cm per second.

so if t=1, v= 2(1) = 2; if t=2, v= 2(2) = 4

Now to calculate the acceleration:

acceleration = $\frac{dv}{dt}$(2t)= 2cm per second squared.

Now apply the same principles to your equations...(Nod)
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