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

Question

anonymous · Answer

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

anonymous · Answer

no, i dont.

anonymous · Answer

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

anonymous · Answer

Ok, so how do I solve this problem?

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

shubhamsrg · Answer

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

anonymous · Answer

Ok.

$$T = 2\pi \sqrt{\frac{4.8}{9.8}}$$

anonymous · Answer

oh it says feet

shubhamsrg · Answer

1 foot= 0.305 meters

anonymous · Answer

we have to use length in feet and time in seconds

anonymous · Answer

is the answer 4.4?