A pendulum bob executing simple harmonic motion has 2cm and 12Hz as amplitude and frequency respectively. Calculate the period of the motion.

qokpara

2 years ago

T=1/f which is =1/12=0.08333=0.08 to one significant figure

Davidemma1234567890

1 year ago

T=1/f
T= Period (s)
f= Frequency (Hz)

Iyke07

2 years ago

what happened to the 2cm?

Davidemma1234567890

1 year ago

The answer is not c but B

Leah300

3 years ago

The answer is c bc T=1\f
I.e 1\12=0.08s
Just neglect the amplitude in the question, don't be confused by it

oluwaseun120

10 years ago

some of the answers are wroung c is the correct answer

babe waterfall

11 years ago

c is correct

adegor25

12 years ago

T- 1/f. 1/12 ....... 0.083 d is correct

Ejcharles

10 years ago

Period = 1/Frequency (Hz) = 1/12 = 0.83333 = 0.08s

adexbob

12 years ago

sir the answer to dis question is 0.83.... cos T= 1/F... where 12 is the frequency which is measured in hz....:. T= 1/12 = 0.8333333

futurestar01

12 years ago

PERIOD (T) = 1/F AND FREQUENCY IS GIVEN AS 12HZ



T = 1/12 = 0.83s



C IS INCORRECT