The frequency, f, in Hz, at which a simple pendulum rocks back and forth is given by 1 , 2 g f π l = where g is the strength of the gravitational field at the location of the pendulum, and l is the length of the pendulum. a. Find the frequency of a pendulum whose length is 1 foot and where the gravitational field is approximately 32 ft/s2 b. The strength of the gravitational field on the moon is about 1 6 as strong as on Earth. Find the frequency of the same pendulum on the moon.

a. 0.9 Hz b. 0.37 Hz

Step-by-step explanation:

The frequency of the simple pendulum f = (1/2π)√g/l where g = acceleration due to gravity and l = length of pendulum

a. Find the frequency of a pendulum whose length is 1 foot and where the gravitational field is approximately 32 ft/s2

To find f on Earth, g = 32 ft/s² and l = 1 ft

So, f =  (1/2π)√(g/l)

f = (1/2π)√(32 ft/s²/1 ft)

f = (1/2π)√(32/s²)

f = (1/2π)(5.66 Hz)

f = 0.9 Hz

b. The strength of the gravitational field on the moon is about 1/6 as strong as on Earth.. Find the frequency of the same pendulum on the moon.

On the moon when acceleration due to gravity g' = g/6,

f =  (1/2π)√(g'/l)

f =  (1/2π)√(g/6l)

f = (1/2π)√[32 ft/s²/(6 × 1 ft)]

f = (1/2π)√(32/s²)/√6

f = (1/2π)(5.66 Hz)/√6

f = 0.9/√6 Hz

f = 0.37 Hz

three-fourths of an hour

36÷48=0.75

0.75=

they are simplified

step-by-step explanation: