# Thread: Plotting graph acceleration against displacement

1. ## Plotting graph acceleration against displacement

A mass of 0.3 kg is suspended from a spring of stiffness 200 N m–1. Ifthe mass is displaced by 10 mm from its equilibrium position andreleased, for the resulting vibration:

Plot a graph of acceleration against displacement (x) (for values of xfrom x = –10 mm to x = +10 mm)

Hugely struggling to plot and draw this graph. Have been reading about a formula to give values for the
acceleration but this formula is giving me numbers which seem way off.

Formula: -(w^2 times displacement)

So, -25.81^2 x 0.1= -66.61561???

Am I on the right track or miles off?? First post and any help is greatly appreciated!

2. ## Re: Plotting graph acceleration against displacement

Originally Posted by james1
A mass of 0.3 kg is suspended from a spring of stiffness 200 N/m. If the mass is displaced by 10 mm from its equilibrium position and released, for the resulting vibration:

Plot a graph of acceleration against displacement for values of x from x = –10 mm to x = +10 mm
$a = -\omega^2 \cdot x = -\dfrac{k}{m} \cdot x = -\dfrac{2000}{3} \cdot x$

note ...

$-0.01 \, m \le x \le 0.01 \, m$

$a$ vs $x$ shown in first attached graph

$a$ vs $t$ shown in second attached graph

3. ## Re: Plotting graph acceleration against displacement

What does "w" mean? And where did you get "-25.81"?

The way I would do this problem is, using "force= mass times acceleration", if the x m from its equilibrium position then the force is -200x = 0.3 x'' (negative since the force is opposite x- if x is downward, the force is upward and vice-versa), where x'' is the second derivative of position- the acceleration. That is the "differential equation" x''+ (200/0.3)x= x''+ (266.67)x= 0 which has the general solution $x= Acos(\sqrt{266.67}t)+ Bsin(\sqrt{266.67}t)= A cos(25.81t)+ Bsin(25.81t)$. We can then calculate that $v= dx/dt= -25.81A sin(25.81t)+ 25.81B cos(26.81t)$.

We are told that the original displacement is 10mm= 1 cm= 0.01 cm (not 0.1) with initial velocity 0 (since it is "released" from that position). That is, we have x(0)= A cos(0)+ B sin(0)= A= -.01 and v(0)= -25.81A sin(0)+ 25.81B cos(0)= 25.81B= 0.

So position x at time t is given by x(t)= -.01 cos(25.81 t). The velocity, as above, is the derivative, 0.2581 sin(25.81 t), and the acceleration is the derivative of that, $(25.81^2)(0.01) cos(25.81t)= 6.66 cos(25.81 t)$.

So the acceleration is $\frac{6.66}{-0.01}= -666= -(25.81)^2$ times x. The graph of that is a straight line, passing through the origin with slope $-(25.81)^2= -666$.

I don't know how much of that makes sense because you don't say what you do understand and what you can do. I suspect that you were given some formulas but you don't say what. You will get a lot better (and simpler) answers next time if you give us all that information to start with.