Velocity and Acceleration

We have seen how we can describe how the displacement of an object moving with shm varies with time. What about its velocity?

There are several ways to approach this, all of which give the same answer.

The blue lines on the first graph below represent the direction and magnitude of the velocity of the object at different times. Imagine a mass on a spring bouncing up and down and think about what direction it is moving in at different times and how fast. You should realise that a graph of velocity against time would be the second graph below.

Clearly this is a cosine. The actual equation for velocity is v = A ω cos ω t. A useful equation is that for the maximum velocity (when cos ω t = 1) and that is vmax = A ω.

You should remember that the gradient of a displacement time graph gives us velocity. Consider how the gradient of the first graph above changes and you should arrive at the second graph.

Finally, for those more mathematically proficient, we can derive the velocity equation by differentiating our equation for displacement.


Consider what the acceleration of the mass below will be in these different positions.

Bear in mind 2 things:
 - the resultant force will be proportional to the displacement ( as F = k x for a spring )
- the acceleration will be proportional to the resultant force ( as F = m a )

 

You should see that the acceleration will be proportional to the displacement but in the opposite direction. Mathematically a is proportional to -y.

So if y is a sine wave then a is a negative sine wave. We can get this result by differentiating our expression for v ( as a = dv/dt ).

We get a = -ω2 A sin ω t.   Now, as y = A sin ω t we can simplify this to a = - ω2 y. So now we have our constant of proportionality.

Below is a chart you should learn which shows the relationship between y, v and a. Now that we can calculate the acceleration of the object we can calculate what forces act upon it at any time or displacement.

Example

A mass of 200g oscillates on a spring with an amplitude of 5cm with a frequency of 2 Hz. Calculate a) its greatest acceleration, b) the largest resultant force acting on it and c) the maximum tension in the spring. At what point will the tension be greatest?

a) a = - ω2 y      so      amax = - ω2 A       now       ω = 2 π f   = 4 π       therefore      amax  = ( 4 π) 2 x 0.05   =   7.89 m/s2

b) F = m a    = 0.2 x 7.89  =  1.58 N

c) The greatest tension will be when the mass is at its lowest position and will equal the resultant force + the weight of the mass i.e. 1.58 + 2 = 3.58N