Question

A meter stick, suspended at one end by a 0.348m long light string, is set into oscillation. The acceleration of gravity is 9.8 m/s^2.
A)Determine the period of oscillation.
Answer in units of s.
B)By what percentage does this differ from a 0.848m long simple pendulum?
Answer in units of percent.

Answer 1

Moment of inertia about suspension point = I about stick middle + mass*(.348+.5)^2

=(1/12) m (1)^2 + m (.719)
= .802 m

T = 2 pi sqrt (I/mgL) = 2 pi sqrt(.802m/(9.8m*.848) )

= 2 pi sqrt(.0965)

= 1.95 seconds

simple pendulum length .848
T = 2 pi sqrt (.848/9.8)
= 1.84 seconds
you can do he percent. Check my arithmetic!