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!

Re: Displacement of a particle

Find the derivative to get the velocity, and the second derivative to find the acceleration with respect to time.

Re: Displacement of a particle

Re: Displacement of a particle

Quote:

Originally Posted by

**srh** Are you sure?

Yes.

$\displaystyle v=d'~\&~a=v'=d''$

Re: Displacement of a particle

Re: Displacement of a particle

Then when is the particle at rest?

Re: Displacement of a particle

$\displaystyle d = 3 sin(40 nt) - 4 cos( 40 nt )$

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

Particle at rest when velocity =0.

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

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

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

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

and then solve for t....

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?

Re: Displacement of a particle

Also the π symbol is supposed to be the pi symbol

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 $\displaystyle ~~\Sigma~~$ that is the LaTeX rap.

[TEX]\pi [/TEX] gives $\displaystyle \pi $.

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.

$\displaystyle velocity = 120n cos(40nt) + 160n sin(40nt)$ --> **which is the derivative of displacement**

$\displaystyle 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:

$\displaystyle velocity = 120(\pi) cos(40\pi(2)) + 160\pi sin(40\pi(2)) = 120\pi$

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

Say the part you don't understand (Hi)

Re: Displacement of a particle

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?

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.

Re: Displacement of a particle

These are equations that describe the motion.

Let's take a simple example

displacement = t for $\displaystyle t\geq0$

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 $\displaystyle velocity = \frac{change\hspace{2mm} in\hspace{2mm} displacement}{change\hspace{2mm} in \hspace{2mm}time}$ = derivative of displacement

from now on I will use *s* to refer to displacement...

velocity = $\displaystyle \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 = $\displaystyle \frac{dv}{dt}$ **(1)**= 0.

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

Second example

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

For the 1st second: displacement = 1 cm

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

we use the formula $\displaystyle velocity = \frac{change \hspace{2mm}in\hspace{2mm} displacement}{change \hspace{2mm}in \hspace{2mm}time}$ = derivative of displacement

from now on I will use *s* to refer to displacement...

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

__Now apply the same principles to your equations...__(Nod)