I have a word problem that I don't know where to begin. I know that I have to use the formula d=9.78t^2. I'm using d as the length and t as the time in seconds (period). The ptoblem is If the length of a pendulum is 60 inches what is the period? Also if the desired period is 3 seconds how long should the pendulum be? Thanks for any help!
word problem in quadratic equation
- Thread starter FMMurphy
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I believe the period of motion for a pendulum is given by the fromula:
\(\displaystyle \L\\T=2{\pi}\sqrt{\frac{L}{g}}\)
\(\displaystyle g=32 ft/sec^{2}\), gravity constant.
L=length of pendulum
Make sure your units are consistent.
As for the cube problem, please do not hijack someone elses post. Kindly start your own thread.
The surface area of a cube is given by \(\displaystyle 6x^{2}\)
Set it equal to your known value and solve for x.
\(\displaystyle \L\\T=2{\pi}\sqrt{\frac{L}{g}}\)
\(\displaystyle g=32 ft/sec^{2}\), gravity constant.
L=length of pendulum
Make sure your units are consistent.
As for the cube problem, please do not hijack someone elses post. Kindly start your own thread.
The surface area of a cube is given by \(\displaystyle 6x^{2}\)
Set it equal to your known value and solve for x.
skeeter
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FMMurphy said:
the formula given ... \(\displaystyle d = 9.78t^2\) ... is close enough for d in inches and t in seconds.
the equation can be solved for t ...
\(\displaystyle d = 9.78t^2\)
\(\displaystyle \frac{d}{9.78} = t^2\)
\(\displaystyle \sqrt{\frac{d}{9.78}} = t\)
now, substitute 60 inches in for d and calculate the value of t, the period.
as far as your last question, substitute 3 sec in for t in the original equation and calculate the value of d.