# Finding the period of a trigonmetic sine curve

• Sep 24th 2009, 11:20 AM
MaverickUK82
Finding the period of a trigonmetic sine curve
Hi Guys!

I am new to this forum, this being my first ever post.

The problem:

I have a problem with calculating the period of a sine curve.
I know the answer to the question - I know how to solve it (or so I thought) but the answer I arrive at doesn't match the one I know (think?!) is right.

I was hoping somebody could guide me through to get the answer I'm looking for?

The question:

Involves finding the period for the function:

h = 35 + 25sin(36t-90)

The h is relative to the height.
The t is relative to time in minutes.

The function is a model of a fairground big-wheel turning.
I have plotted this on a graph and have found the time it takes for the big wheel to do one revolution is 10 minutes.

I thought the formula to calculate the period from the above was:

T=((2*pi)/36) = 0.174533

But how can that be the period when I know from my graphs etc that it takes 10 minutes to do one revolution?

I then thought the 0.174533 might be the frequency, but the reciprocal of the frequency should be period (I'm right on that - right?), but the reciprocal of 0.174533 is 5.73.

I have been on this for over 2 days now, almost non-stop and can't fathom it out. The closest I can get is by multiplying my answer 0.174533 by 60 to get 10.47.

• Sep 24th 2009, 01:19 PM
Amer
Quote:

Originally Posted by MaverickUK82
Hi Guys!

I am new to this forum, this being my first ever post.

The problem:

I have a problem with calculating the period of a sine curve.
I know the answer to the question - I know how to solve it (or so I thought) but the answer I arrive at doesn't match the one I know (think?!) is right.

I was hoping somebody could guide me through to get the answer I'm looking for?

The question:

Involves finding the period for the function:

h = 35 + 25sin(36t-90)

The h is relative to the height.
The t is relative to time in minutes.

The function is a model of a fairground big-wheel turning.
I have plotted this on a graph and have found the time it takes for the big wheel to do one revolution is 10 minutes.

I thought the formula to calculate the period from the above was:

T=((2*pi)/36) = 0.174533

But how can that be the period when I know from my graphs etc that it takes 10 minutes to do one revolution?

I then thought the 0.174533 might be the frequency, but the reciprocal of the frequency should be period (I'm right on that - right?), but the reciprocal of 0.174533 is 5.73.

I have been on this for over 2 days now, almost non-stop and can't fathom it out. The closest I can get is by multiplying my answer 0.174533 by 60 to get 10.47.

it is the same how you find t=10 min anyway

$\frac{2\pi}{36 }$ pi here not 22/7 pi equal 180 degree that's you fault so

$\frac{2\pi}{36 } =\frac{360}{36} = 10$
• Sep 24th 2009, 01:43 PM
MaverickUK82
Brilliant... so it was working in radians and just needed to be in degrees! Excellent!

I don't however, understand your comment:
"pi here not 22/7 pi equal 180 degree that's you fault so".

where did the 22/7 pi come from?
• Sep 25th 2009, 03:29 AM
Amer
Quote:

Originally Posted by MaverickUK82
Brilliant... so it was working in radians and just needed to be in degrees! Excellent!

I don't however, understand your comment:
"pi here not 22/7 pi equal 180 degree that's you fault so".

where did the 22/7 pi come from?

I mean when you find the period in this

Quote:

Originally Posted by MaverickUK82
I thought the formula to calculate the period from the above was:

T=((2*pi)/36) = 0.174533

you sub pi value 22/7 but you should sub 180 instead of 22/7
• Sep 25th 2009, 05:23 AM
pacman
pi is approximately equal to 22/7, BUT we are converting from RADIANS measure to DEGREES measure, MHF Amer is right in his comment . . . that is 2 pi rad = 360 degrees.
• Sep 25th 2009, 03:27 PM
MaverickUK82
Thanks guys - get it now - reps to both of you!