amplitude, perodic time, angular velocity, phase angle and lead time!

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December 2nd 2010, 11:29 AM
maths99

amplitude, perodic time, angular velocity, phase angle and lead time!

i am working through a question, and just wanted to make sure i am on the right lines. so guidence and help be great the question is as follows:

an oscillating mechanism has a maximum displacement, d of 1.6m and a freqency of 20Hz. At a time t=0 sec the displacement is +50 cm.

determine the amplitude, periodic time, angular velocity, phase angle and lead time, then express the displacement in the form Rsin (vt + a)

my working:

amplitude A = maximum displacement
A = 1.6m

angular velocity: omega symbol = 2pf
=2pf 20Hz
= 40prad/sec
phase angle, x use displacement, d = A sin (v t + a)
d = 1.6sin (40pt+a)

when t=0, d =0.50m
0.50=1.6sin(40p(0)+a)
0.50=1.6sin(a)
0.50/1.6=sin a
sin-1 0.50/1.6 = a
=18.2 degress

18.2 x p rads/180

a=18.2 x p rads/180

a= 0.317rad

displacement = 1.6sin(40p (t) + 0.317)
=1.6 (40t+0.317)

this is where i get bit lost, to get to the lead time. i know the formula a/v
any help to get there by great thanks
December 2nd 2010, 03:49 PM
pflo

Quote:

Originally Posted by maths99

i am working through a question, and just wanted to make sure i am on the right lines. so guidence and help be great the question is as follows:
an oscillating mechanism has a maximum displacement, d of 1.6m and a freqency of 20Hz. At a time t=0 sec the displacement is +50 cm.
determine the amplitude, periodic time, angular velocity, phase angle and lead time, then express the displacement in the form Rsin (vt + a)

In general, you can talk about any function as a transformation of a parent function. The parent function for all sine functions is $d=sin(t)$ (or whatever you want to call the variables - I've used d for displacement and t for time). This parent function has an amplitude of 1 and a period of $2\pi$ .

A transformation would look like this: $d=Asin[b(t+c)]+u$ , where A is the amplitude (how much the graph is stretched or squshed vertically), b is a factor related to the frequency or period (it squishes or stretches the curve in a horizontal direction on the graph), c is the phase shift (which effects where the graph intersects the vertical axis - when t=0) and u is the vertical shift.

In your case, the amplitude is 1.6.

Since f=20 hertz, a complete cycle takes 1/20 of a second. Since the sine function has a natural period of $2\pi$ , it would need to be squished by a factor of $40\pi$ (so this is 'b' in my version of the formula the formula).

Regarding the phase shift, this is the amount the parent graph should be moved to meet the initial conditions. So, $0.5=1.6sin[40\pi(0+b)]$ . Solving this equation gives $40\pi b = 0.3178$ . In your equation, you have distributed the $40\pi$ through the $40\pi (t+b)$ , so this is the equivalent of what you got as well. However, from what you describe there is no way of knowing whether the displacement is increasing or decreasing at time t=0. It could be doing either. Your equation is correct if the displacement is increasing. But if the displacement is decreasing when t=0, then you would have to move the graph further. The actual phase shift of your solution is $\frac{0.3178}{40\pi}=0.002529$ . If the displacement were decreasing, the displacement would be half a cycle minus this amount, or $1/40-0.002529=0.02247$ . Multiplying to get this in the form or your equation would give you a phase angle of $40\pi * 0.02247=2.8238$ .

So there are two equations that will satisfy these conditions:
$d=1.6sin(40\pi t+0.3178)$ and
$d=1.6sin(40\pi t+2.8238)$ .

Overall a good job by you!!!

If the formula for 'lead time' is $\frac{a}{v}$ , then you should just plug in those numbers. This is the equivalent of the 'phase shift' in my formula: $\frac{0.3178}{40\pi}=0.002529$ or $\frac{2.8238}{40\pi}=0.02247$ . Either way, it is 0.002529 seconds away from its center of displacement (frequently called the equilibrium position), but it would pass this position before t=0 (and headed in the positive direction) in the first case and after t=0 (and heading in the negative direction) in the second case.
December 3rd 2010, 06:45 AM
maths99

thanks for your help, i was on the right lines. now finished.
after completed the above, college has asked if we could look at this, slightly confused at what they are asking.

There are two occasions in a cycle when the displacement is +50cm

If the second displacement is +50cm occurred at t=0

Calculate phase angle

Sketch the graph of one cycle of displacement against time.
December 6th 2010, 02:52 AM
pflo

1 Attachment(s)

Quote:

Originally Posted by maths99

thanks for your help, i was on the right lines. now finished.
after completed the above, college has asked if we could look at this, slightly confused at what they are asking.

There are two occasions in a cycle when the displacement is +50cm

If the second displacement is +50cm occurred at t=0

Calculate phase angle

Sketch the graph of one cycle of displacement against time.

A sine curve goes out, then comes back in. In this case, the displacement increases from the equilibrium point (to +0.50 and past) then it decreases (back through +0.50) toward the equilibrium point. After the dsplacement goes the other direction (the negative) it comes back to repeat the process again. So the displacement is indeed +0.50 two different times, once as the displacement is increasing and once as it is decreasing. Check out the graph I'm attaching. Note, however, that this is the graph of $y=sin(x)$ , which is the partent function to your equation. The parent function starts at the equilibrium point when t=0 (this graph also has the x-axis labeled with radians, not time).Attachment 19984

There are at least two ways to think about how to arrive at the equation you need for this new quesiton. But first, realize that one difference between your original function and the the parent function is shifted horizontally. The point (0.002529, 0.50)on the parent graph has moved over so that it now sits on the y-axis and your graph has the point (0,0.50). In effect the graph has moved 0.002529 to the right.

The new question involves when the displacement will be at +0.50 again. This point is symmetric around the maximum, which is 1/4 of a cycle from the origin of the parent function (or 0.0125 seconds). The first time the displacement is at +0.50 is 0.009971 seconds before it reaches the maximum, so it will be at +0.50 again the same time afterwards - at 0.02247 seconds. So the parent function should get moved that far. Another way to look at it is that it reaches +0.50 displacement 0.0125 seconds after it leaves the equilimbium point and it will take the same amount of time to reach the equilibrium point from 0.50 on the way back. So it will hit a half-cycle 0.002529 seconds after it is at +0.50. The half-cycle is 1/40 of a second, so the parent function would have to move over $\frac{1}{40}-0.002529=0.02247$ seconds.

The figure - 0.02247 - is the 'c' in the equation: $d=Asin[b(t+c)]+u$ and, when the $40\pi$ is distributed it gives you the second answer I originally posted: $d=1.6sin(40\pi t+2.8238)$