Pendulum Problem using derivative of sine/cosine

**Pendulum completes swing (back and forth) each 6s. As it swings, the distance, d, in cm, depends on time t, in seconds. At t=1.3 d is at its max of 110cm. Its min is 50cm. D is sinusoidal function of t. **

I got the equation y=80+60cospi/3(t-1.3) using the methods from the book (find the sine axis, amplitude, high point and B), is there a better way/ faster way to do it?

The second question asks for the fastest pendulum swing, and where it is swinging its fastest. I don't understand how to solve for this question.

For the third question what is the first positive value of t at which the pendulum is swinging 0cm/s and where is the pendulum at this time- do you find the third derivative y'' and then solve for y''=0? then put the value t into the main equation y?

Re: Pendulum Problem using derivative of sine/cosine

You will need to find the formula for speed. Then maximize that function.

For the third question, set the speed function to 0 and solve for the minimum positive value of t.

For reference, the speed function is the magnitude of the velocity function.

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Re: Pendulum Problem using derivative of sine/cosine

$\displaystyle T=2\pi \sqrt{l/g}=6$ -> L = 894.26 cm (length of pendulum), g = 980.665 cm/s/s

$\displaystyle \theta (\text{t})\text{=}\text{c1}* \text{Cos}\left[\frac{\sqrt{g} t}{\sqrt{l}}\right]+\text{c2}* \text{Sin}\left[\frac{\sqrt{g} t}{\sqrt{l}}\right]$

c1=0.0069762

c2=0.03282

theta = angle of pendulum with vertical

http://mathhelpforum.com/attachment....1&d=1349762444

max velocity = 31.421 cm/s at theta = 0 (vertical position)

1st velocity =0 at t=1.3

http://mathhelpforum.com/attachment....1&d=1349763682

You can easily transfer theta(t) to horizontal d(t).

Re: Pendulum Problem using derivative of sine/cosine

Can you clarify the formula with c1 and c2?

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Re: Pendulum Problem using derivative of sine/cosine