A particle moves in a circle in such a way that the x- and y-coordinates of its motion are given in meters as functions of time t in seconds by:
x=5cos(3t)
y=5sin(3t)
What is the period of revolution of the particle?
Here's what my train of thought was... dy/dt over dx/dt would equal dy/dx which equals velocity. So velocity would be -cot(3t). But then how do I find r? Plus none of the answers here have a trig function in them... ???????
I was looking for r because period = [2(pi)(r)]/v......
And how did you get to 3t=2Pi*n???? What equation is that?????
solve t by the method I gave, for get r.
3t=0, 2PI, 4PI....is not the period of a sin function =2PI or a multiple thereof?
Well, I am 14 years late to answer this, but just hear me out!
We can see that the x component of the co-ordinate is x=2 cos(3t)
Again, recall that in a circular motion the co-ordinates can be written as:
A cos wt (for x co-ordinate) {Here w is omega, the angular velocity}
See the relation? A cos wt <=> 2 cos 3t?
omega=3
Also, omega = 2*pi*frequency
Or omega=(2*pi)/Time period
3=2pi/T
T=2/3 pi or 2*pi/3
Sorry, I meant 5 cos wt and not 2 cos wt
Please ignore that
To find the period of revolution of the particle, we need to understand the motion of the particle in a circle.
The equation x = 5cos(3t) represents the x-coordinate of the particle as a function of time, and the equation y = 5sin(3t) represents the y-coordinate of the particle as a function of time.
We can see that these equations are in the form of a parametric representation of motion in a circle, where x and y both vary sinusoidally with time. The coefficient 3 in both equations indicates that the particle completes one full revolution in (2π/3) seconds.
Now, let's find the period of revolution.
The period of a sinusoidal function is the time it takes for the function to complete one full cycle. In this case, the period of revolution represents the time it takes for the particle to complete one full circular motion.
To find the period, we need to find the time it takes for both x and y to return to their initial values. Since both x and y vary with time in the same manner, they will complete one full cycle simultaneously.
To determine the period, we need to find the time when the argument of the trigonometric functions completes one full cycle. In this case, the argument of both cosine and sine functions is 3t.
To find the period, we set the argument equal to 2π (one full cycle), and solve for t:
3t = 2π
t = (2π/3)
Therefore, the period of revolution of the particle is (2π/3) seconds.
Note that the velocity of the particle (-cot(3t)) is not necessary to find the period of revolution. The period can be determined solely by examining the coefficients and arguments in the given equations.
Hmmmm. You are not asked for r, you are asked for period, in time.
3t= 2PI*n n=0, 1, ...
t= 2/3 PI n try n=1 for the first rev.