Can anyone help me with this question please?
(Image removed from quote.)
Thanks!
We know that the displacement function will look something like
=A\cos(kt))
We are given the amplitude (5cm), and the period. Recall that
This is a formula that you just need to memorise! So, we solve for k to find that
This, our equation of motion looks like
=5\cos(\frac{\pi}{3}t))
I have omitted any phase shift or 'up/down' shift, as the question doesn't give you initial values that you could use to solve for them. The question also doesn't ask for anything that could depend on them, so the simple above function works just fine. Now, the velocity when the particle is 2.5m from the centre of motion is the velocity when the particle is
at an out limit of it's motion. The centre of motion must be 2.5cm (half of the amplitude) away from each end point. So, a) is asking for the velocity at the end-point, which will be zero (as the particle is 'turning around', and thus is standing still for just a moment).
The maximum acceleration, however, will occur at this outermost point (ie. where velocity equals zero). To find the maximum acceleration, we differentiate twice to find acceleration.
=-5(\frac{\pi}{3})^2\cos(\frac{\pi}{3}t))
Now, note that trigonometric functions oscillate between -1 and 1. Thus, the maximum value for the double derivative of displacement will occur when the trig function is equal to -1 (not one, as the coefficient is negative). So, we get
=a_{max}=-5(\frac{\pi}{3})^2)
Home
Help
Search
Login
