Can somebody help please??
The height h in cm of a bicycle pedal above the ground at time t seconds is given by
h = 30 + 15 cos 90 t
How do I find the number of revolutions in one minute?
Thanks
The height is a periodic function of $\displaystyle t$, as it should be. The frequencyOriginally Posted by radius
(cycles per second) of the pedal is the frequency of the $\displaystyle \cos (90.t)$
part of the height formula. A cosine based signal of frequency $\displaystyle f$ is:
$\displaystyle k. \cos (2 \pi.f.t)$, so in this case:
$\displaystyle 90=2 \pi.f $.
So
$\displaystyle f=45/ \pi$ cycles per second,
or
$\displaystyle \frac{45\times 60}{\pi}$ cycles per minute.
RonL
Here is one way.Originally Posted by radius
h = 30 +15cos(90t)
is a cosine curve whose:
---"original" or "basic" origin is shifted to 30cm above the (0,0) of the t,h-axes setup. There is no horizontal shift.
Its amplitude is 15cm. That means at maximum height, the pedal is (30+15) = 45cm above ground, and at minimum height, the pedal is (30-15) = 15cm above the ground. Very reasonable distances from the ground.
Its period is 360/90 = 4sec per cycle. One period = one cycle = one revolution of the pedal.
So one revolution is 4sec. In one minute, that will be 60/4 = 15 revs per minute.
Therefore, the pedal moves at 15 rpm. --------answer.
Reasonable speed.
Let us check that.
At t = 0 sec----h = 30 +15cos(90*0) = 30 +15cos(0) = 30 +15(1) = 45 cm.
At t = 1 sec----h = 30 +15cos(90*1) = 30 +15cos(90deg) = 30 +15(0) = 30 cm.
At t = 2 sec----h = 30 +15cos(90*2) = 30 +15cos(180deg) = 30 +15(-1) = 15 cm.
At t = 3 sec----h = 30 +15cos(90*3) = 30 +15cos(270deg) = 30 +15(0) = 30 cm.
At t = 4 sec----h = 30 +15cos(90*4) = 30 +15cos(360deg) = 30 +15(1) = 45 cm.
See, at 4sec, the pedal is back at the initial 45cm above ground. That is 4sec per revolution. In one minute, that is 15 revs. That is 15 rpm.