1. ## simple harmonic motion

Ok, I am studying for a test and working through some practice problems on a handout from teacher. The problem is, I am getting a different answer than the one he gives for this application problem:

"A tractor tire has a diameter of 6 feet and is revolving at a rate of 45 rpm. At t=0 a certain point is at height 0.
A) Write an equation with phase shift 0 to describe the height of the point above the ground after 't' seconds.
B) How high is the point above the ground after 28 seconds?"

I get this:
h = -3cos(t*8pi/3) + 3
and at t = 28, h = 4.5

but the answer given is this:
h = -3cos(t*3pi/2)+3
and at t = 28, h = 0

The difference is the value for k, which I got by converting the period from revs per min (45/1) to rev per second (45/60 or 3/4), and then k = 2pi(4/3). The supplied answer inverts this, and I am wondering if this is a mistake or if I am looking at it incorrectly. Thanks.

2. Originally Posted by sinewave85
Ok, I am studying for a test and working through some practice problems on a handout from teacher. The problem is, I am getting a different answer than the one he gives for this application problem:

"A tractor tire has a diameter of 6 feet and is revolving at a rate of 45 rpm. At t=0 a certain point is at height 0.
A) Write an equation with phase shift 0 to describe the height of the point above the ground after 't' seconds.
B) How high is the point above the ground after 28 seconds?"

I get this:
h = -3cos(t*8pi/3) + 3
and at t = 28, h = 4.5

but the answer given is this:
h = -3cos(t*3pi/2)+3
and at t = 28, h = 0

The difference is the value for k, which I got by converting the period from revs per min (45/1) to rev per second (45/60 or 3/4), and then k = 2pi(4/3). The supplied answer inverts this, and I am wondering if this is a mistake or if I am looking at it incorrectly. Thanks.
Here is one way.

The locus of the point in question relative to t in seconds is that of a sinusoidal curve---sine or cosine curve. Since the h is 0 at t=0, and the "spoke" or "center" of the tire is 3 ft above the ground, then the cosine curve is more applicable here because at h=0, the graph is at its lowest point.

The equation then has the form
h = A*cos[B(t-C)] +D --------------(i)
where
A = amplitude = 3 ft here.
B = frequency in revolution/sec ----[Period = 2pi/B]
C = horizontal shift or phase = 0 here
D = vertical shift = 3 ft above the ground here.

Hence, (i) becomes
h = 3cos[B*t] +3 --------------(ii)

Only B is missing.
The tire moves at 45 rpm, at 45 revolutions per minute. Per second, that is
(45 rev/min)(1min/60sec) = (45/60) rev/sec = 3rev /4sec.
That means frequency B = 3rev/4sec. In radians, since 1 rev = 2pi radians,
Thus, (ii) now becomes
h = 3cos[(3pi/2)*t] +3 -------------------(iii)

It is given that at t=0 sec, the h is 0 ft, so, in (iii),
0 = 3cos[(3pi/2)*t] +3
Without going any further, just by looking, it is obvious that for the righthand side to be equal to zero, "3cos[(3pi/2)t]" must be negative.
Thererfore, finally, the (iii) becomes
h = -3cos[(3pi/2)t] +3 ---------------------the equation, answer.

---------------------------
B) "How high is the point above the ground after 28 seconds?"

h = -3cos[(3pi/2)(28)] +3