Velocity, Limits: position of pendulum t seconds after....

[alicia3138]

i can't figure out this problem:

The position of a pendulum t seconds after it is first set in motion is given by the function \(\displaystyle s(t) = 5 cos(\frac{4{\pi}t}{3})\), where s(t) measures how far from its center line the pendulum's bob is, positive measurements indicating that the bob is to the right of the centerline, negative measurements indicating that the bob is to the left of the centerline, and a measurement of zero indicating that the bob is directly over the centerline.

(b) Estimate the velocity of the bob at t = 5 seconds.

Answer: About -18 units per second.

i email my prof and he told me:

Approximate the velocity of the bob at t=5 using an average velocity over a time interval like [5,6] or [5,5.5] or [5,5.5], etc. The average velocity over [5,6] would be given by (s(6)-s(5))/(6-5)

but when i do that calculation on top i get like 7.5 or something. what am i doing wrong?

 

[skeeter]

Re: Velocity, Limits

if your position function is \(\displaystyle s(t) = 5\cos \left(\frac{4\pi}{3} t \right)\), then your calculation is correct.

\(\displaystyle s(6) - s(5) = 5\cos(8\pi) - 5\cos\left(\frac{20\pi}{3}\right) = 5 - 5\left(-\frac{1}{2}\right) = 7.5\)

to get a good estimate for v(5), you'll have to use a much smaller interval of time ... try \(\displaystyle \frac{s(5.05) - s(4.95)}{5.05 - 4.95}\) ... see what you get.

 

[galactus]

If we take the derivative of s(t) we get \(\displaystyle s'(t)=v(t)=\frac{-20sin(\frac{4t}{3})}{3}\)

At t=5, we get -2.49

 

[skeeter]

If we take the derivative of s(t) we get \(\displaystyle s'(t)=v(t)=\frac{-20sin(\frac{4t}{3})}{3}\)

At t=5, we get -2.49

"g" ... I think pi was left out of the original post in the argument for cosine.

 

[galactus]

Yes indeed. I should've caught that. We do get -18 when plugging t=5 into the derivative.