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A meter stick is attached to one end of a rigid rod with negligible mass of length l = 0.302 m. The other end of the light rod is suspended from a pivot point, as shown in the figure below. The entire system is pulled to a small angle and released from rest. It then begins to oscillate. A meter stick hung from a rod of length l. The rod is attached to the ceiling. The rod and meter stick extend downward in a straight line making a small angle with the vertical. (a) What is the period of oscillation of the system (in s)? (Round your answer to at least three decimal places.)

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ajinems

The period of oscillation of the system nearest to three decimal places

= 1.092 seconds

The period of an oscillation occurring in a system is the time taken to complete one cycle.

The formula that is used to calculate the period of oscillation (T) is

                = 2π√[tex]\frac{l}{g}[/tex]

But,

π = 3.14159 (constant)

g= 10m/s² (acceleration due to gravity)

l = 0.302 m

Therefore T = 2 × 3.14159 × √[tex]\frac{0.302}{10}[/tex]

                    = 6.28318 x √0.0302

                    = 6.28318 x 0.17378

                    = 1.09189s

                    = 1.092 seconds ( to the nearest three decimal places)

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