Question

Clock #2: A clock has a radius of 6 inches. The center is 14.5 inches below the ceiling. Record the distance as you did for Clock #1 and write a sine or cosine equation.

Answer 1

Answer:

The question is incomplete, I assume that we want to write an equation that defines the position (or distance) between the tip of the second hand of the clock and the ceiling.

We know that:

The center of the clock is 14.5 in below the ceiling.

The radius of the clock is 6 inches.

So, a general cosine equation is written as:

f(x) = A*sin(w*t) + M

where:

A = amplitude, in this case, is equal to the radius of the clock

w = angular frequency

t = time in seconds

M = midline (the oscillation is around this point), in this case, would be the distance between the center of the clock and the ceiling, because the clock is 14.5 inches below the ceiling, we can write this as M = -14.5 in

Replacing these two in the equation we have:

f(t) = 6in*cos(w*t) - 14,5in

Now, we know that at t = 0s the tip of the second hand of the clock is in its most high point, so at t = 0 we have the maximum, this is why we used a cosine function, because the maximum of the cosine function is at:

cos(0) = 1

We also know that the minimum will be at t = 30 seconds (when the tip of the second hand is in the bottom part of the clock)

then we need to have:

cos(w*30s) = -1

Remember that:

cos(pi) = -1

then:

w*30s = pi

solving for w, we get:

w = pi/30s

Then the equation for the distance between the tip of the second hand of the clock and the ceiling is:

f(t) = 6in*cos(t*pi/30s) - 14.5 in