1. ## Sinusoidal Functions

12/15

I have the following math problem:

Kala is jumping rope, and the rope touches the ground every time she jumps. She jumps at the rate of 40 jumps per minute, and the distance from the ground to the midpoint of the rope at its highest point is 5 feet. At t=0, the height of the midpoint is zero. Write a function with phase shift 0 for the height of the midpoint of the rope above the ground after t seconds.

I am not sure whether to use sine or cosine for the function.

I don't know where to begin on this problem. Could someone help? Thanks.

Joanie

2. ## Sinusoidal Functions

Hi Joanie -

Originally Posted by Joanie
Kala is jumping rope, and the rope touches the ground every time she jumps. She jumps at the rate of 40 jumps per minute, and the distance from the ground to the midpoint of the rope at its highest point is 5 feet. At t=0, the height of the midpoint is zero. Write a function with phase shift 0 for the height of the midpoint of the rope above the ground after t seconds.
General sinusoidal functions will look something like

$h=a \pm b\sin(\omega t+\alpha)$ or $h=a \pm b \cos(\omega t + \alpha)$

In these functions:

• the mean value of $h$ is $a$
• the amplitude is $b$
• the period is $\frac{2\pi}{\omega}$
• the phase shift is $\alpha$

We need the phase shift to be zero, so $\alpha = 0$.

So the function will be of the form:

$h=a \pm b\sin(\omega t)$ or $h=a \pm b \cos(\omega t)$

But which one do we choose? Well, this will be determined by the value of $h$ when $t=0$. So we need look at the values of $\sin 0$ and $\cos 0$

Now $\sin 0 = 0$ and $\cos 0 = 1$

So the possible values of $h$ when $t = 0$ are:

• $a$
• $a+b$
• $a-b$

If $h=a$ when $t=0$, use $h=a \pm \sin(\omega t)$ (Use the + sign if $h$ is increasing when $t=0$, and the - sign if it's decreasing.)

If $h=a+b$ when $t=0$, use $h=a+b \cos(\omega t)$

If $h=a-b$ when $t=0$, use $h=a-b \cos(\omega t)$

Can you see what the values of $a$ and $b$ are? Can you see which of the four possible options to use? Can you see how to work out the period, and hence the value of $\omega$?

Let me know if you need more help.

12/19

I have read your explanation of my word problem. The formula I am using:

Asin(or cos)[pi/k +(or)-c/k] + h

I have played around with the info in the word problem and have come up with a few different equations. I do not know if they are correct. Here they are:

h = 2.5 cos[3pi(t)] + 2.5

h= -2.5cos[4pi/3(t)] + 2.5

h= 2.5sin[3pi(t)] + 2.5

h = -2.5sin[4pi/3(t)] + 2.5

I am still not sure whether to use sine or cosine.
These equations are very similar--however, with all the other problems I have done similar to this one, I have used sine.

Thanks for your help. Hope to hear from you soon.

Joanie

4. ## Sine or Cosine

Hello Joanie -

Originally Posted by Joanie
12/19

I have read your explanation of my word problem. The formula I am using:

Asin(or cos)[pi/k +(or)-c/k] + h

I have played around with the info in the word problem and have come up with a few different equations. I do not know if they are correct. Here they are:

h = 2.5 cos[3pi(t)] + 2.5

h= -2.5cos[4pi/3(t)] + 2.5

h= 2.5sin[3pi(t)] + 2.5

h = -2.5sin[4pi/3(t)] + 2.5

I am still not sure whether to use sine or cosine.
These equations are very similar--however, with all the other problems I have done similar to this one, I have used sine.

Thanks for your help. Hope to hear from you soon.

Joanie
You are right to say $a =2.5$ and $b=2.5$ (to use the letters I used in my first posting). So we have equations that look like:

$h=2.5 \pm 2.5\sin(\omega t)$ (1)

and

$h=2.5 \pm 2.5 \cos(\omega t)$ (2)

And your first question is: Which one do I choose, sine or cosine? Well, look at the values of sine and cosine when $t=0$. $\sin(0) = 0$, so if we use either of the equations numbered (1), we should get the value $h=2.5$ when $t=0$. Is that what we want?

No, because we're told that $h=0$ when $t=0$.
At t=0, the height of the midpoint is zero.
So let's try the equations numbered (2). (There are two equations here: one taking the + sign part of $\pm$, and the other taking the - sign.) These equations use the value of cosine when $t=0$, and the value of $\cos(0)$ is $1$.

Let's try the + sign first:

$h=2.5+2.5 \cos(\omega t)$

When $t = 0$, $h=2.5 + 2.5\times 1=5$

That's not right either. We want $h=0$.

So finally, try:

$h=2.5-2.5\cos(\omega t)$

When $t=0$, $h = 2.5 - 2.5\times 1 = 0$

Correct!

So the equation we want will look like:

$h=2.5-2.5\cos(\omega t)$

Finally, what about the value of $\omega$? Well, we know that functions like this have a period (the time it takes for a complete cycle) of $\frac{2\pi}{\omega}$. So what is the period of our rope? Well, it takes 60 seconds to complete 40 turns, so the time it takes to complete 1 turn is $\frac{60}{40} = 1.5$ seconds.

So: $\frac{2\pi}{\omega}=1.5$

$\implies 2\pi = 1.5 \times \omega$

$\implies \omega = \frac{4\pi}{3}$

So the equation we want is:

$h=2.5-2.5 \cos (\frac{4\pi}{3}t)$