A pendulum has a frequency of 2 hz (frequency =1/ time period). Find its length in feet. Round to the nearest hundredth of a foot. T=2pie(l/32)^1/2

Question

This thing is hard

Answer 1

Actually, it's not that bad. 

I'm going to use metric units, and at the end will convert to ft.

 

L = Length

g = gravity (9.81 m/s²)  (@ surface of earth)

f  = freq = 2 Hz

 

if freq = 2 Hz,  then period P = 1/f = 0.5 sec

 

Period (pendulum) = 2*pi*sqrt(L/g)

now, just solve for L (length)

 

P / (2*PI) = SQRT(L/g)      now square both sides

 

(P / (2*PI))² = L/g            mult by g both sides

 

g*(P / (2*PI) )² = L           now just plug in values for P, g

 

9.81*(0.5 / (2*pi) )² = L 

 

L = 0.0621 meters = 6.2122 cm 

 

now convert to feet.  30.48 cm/ft so 6.21 cm/ 30.48 cm/ft

 

L = 0.2038 FT     to nearest 0.01    L = 0.20 FT