Question

Find the period, to the nearest tenth of a second, of a pendulum of length 4.8 feet.

Answer 1

Do you know the formula for simple harmonic oscillation of a pendulum's period?

Answer 2

no, i dont.

Answer 3

Well, it's 2*pi*sqrt(L/g) where L = length of pendulum and g = gravity

Answer 4

Ok, so how do I solve this problem?

Answer 5

Just plug in the numbers...which is only the length of the pendulum

Answer 6

is this correct 2*pi*sqrt(4.8) ?

Answer 7

you have to divide the length by the force of gravity first (-9.8)

Answer 8

i don't really get it. I'm not really familiar with the formula

Answer 9

and 4.8 wont be it..you'll have to convert it to meters ..since you're usong g=9.8 m/s

Answer 10

Ok.

\[T = 2\pi \sqrt{\frac{4.8}{9.8}}\]

Answer 11

oh it says feet

Answer 12

1 foot= 0.305 meters

Answer 13

we have to use length in feet and time in seconds

Answer 14

is the answer 4.4?