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aced3f3atsall · Mar 26, 2012, 02:30 AM

Verify the identity cot (theta-pi/2)=-tan theta

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aced3f3atsall · Mar 26, 2012, 02:37 AM

The equation h=7cos (pi/3t) models the height h in centimeters after t seconds of a weight attached to the end of a spring that has been stretched and then released.
a. Solve the equation for t.
b. Find the times at which the weight is first at a height of 1 cm, of 3 cm, and of 5 cm above the rest position. Round your answers to the nearest hundredth.
c. Find the times at which the weight is at a height of 1 cm, of 3 cm, and of 5 cm below the rest position for the second time. Round your answers to the nearest hundredth.

ebaines · Mar 26, 2012, 06:02 AM

Originally Posted by aced3f3atsall

Verify the identity cot (theta-pi/2)=-tan theta

Replace the cotangent function with sin/cos, and then apply the identiies for sin(a-b) and cos(a-b), where a = theta and b = pi/2.

ebaines · Mar 26, 2012, 06:04 AM

Originally Posted by aced3f3atsall

The equation h=7cos (pi/3t) models the height h in centimeters after t seconds of a weight attached to the end of a spring that has been stretched and then released.
a. Solve the equation for t.
b. Find the times at which the weight is first at a height of 1 cm, of 3 cm, and of 5 cm above the rest position. Round your answers to the nearest hundredth.
c. Find the times at which the weight is at a height of 1 cm, of 3 cm, and of 5 cm below the rest position for the second time. Round your answers to the nearest hundredth.

Please show us how you have attempted to solve this, and where you have gotten stuck, and then we'll help you along. But here's a hint: you will need to apply the Arccos function to both sides, and recall that Arccos(cos(x)) = x.

prayer62 · May 2, 2012, 07:23 AM

a. t=(3arccos(h/7)/pi)
b. t=4.28/pi for 1cm, t=3.38 for 3cm, t=2.33 for 5cm

I got this so far but got stuck on c.

ebaines · May 2, 2012, 08:30 AM

Originally Posted by prayer62

a. t=(3arccos(h/7)/pi)
b. t=4.28/pi for 1cm, t=3.38 for 3cm, t=2.33 for 5cm

I got this so far but got stuck on c.

The weight will reach these displacement for the second time as it rebounds from maximum negative extension. So first find the time to max negative displacement, then add the time interval(s) from that point to these displacements. Couple of hints:

a. The time to go from max negative didplacement to position -x is the same as to go from max positive displacement to position +x.
b. The time to reach max negative displacement is 1/2 the period T of this sinusoisdal motion. For motion of the form $x(t) = A \cos(\omega t) , \ \omega$ is the radial velocity in radians/second and the period T for the sinusoidal motion is $T = \frac {2 \pi} {\omega}$

prayer62 · May 2, 2012, 12:27 PM

Okay thanks!

prayer62 · May 2, 2012, 12:38 PM

So I got.
c. t=2.14 for 1cm, t= 1.67 for 3cm, t=1.17 for 5cm
would this be correct?

ebaines · May 2, 2012, 01:26 PM

Originally Posted by prayer62

So I got.
c. t=2.14 for 1cm, t= 1.67 for 3cm, t=1.17 for 5cm
would this be correct?

No. The time it takes for the weight to reach the max negative displacement is T/2, so you know the answers must be greater than that. What did you get for T?

prayer62 · May 2, 2012, 01:45 PM

I used my answer for a. t=(3arccos(h/7)/pi)

ebaines · May 2, 2012, 02:28 PM

Originally Posted by prayer62

I used my answer for a. t=(3arccos(h/7)/pi)

What I meant is: what do you calculate for the period of the sinusoidal motion T (capital T)? Hint - it's the twice the value of time for when h = -7. The graph below may help you visualize this.