The length of a simple pendulum is 0.78 m and the mass of the particle (the \'\'

Question

The length of a simple pendulum is 0.78 m and the mass of the particle (the ''bob'') at the end of the cable is 0.25 kg. The pendulum is pulled away from its equilibrium position by an angle of 8.15 degree and released from rest. Assume that friction can be neglected and that the resulting oscillatory motion is simple harmonic motion. (a) What is the angular frequency of the motion? ______________ rad/s (b) Using the position of the bob at its lowest point as the reference level, determine the total mechanical energy of the pendulum as it swings back and forth. ___________ J (c) What is the bob?s speed as it passes through the lowest point of the swing? ________________ m/s

Explanation / Answer

The Time period of the pendulum

       T = 2? ? [ L/g ]

          = 2? ?[0.78 /9.8 ]

          =1.77 s

The angular frequency is

        ?= 2? / T

            =2 ? / 1.77

           = 3.55 rad/ s

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The height through which the bob is lifted up is

        h = L ( 1 - cos? )

           =0.78 ( 1 - cos 8.1.5 )

           =7.88x10-3 m

The total mechanical energy is

        E = m g h

           =0.25(9.8)7.88x10-3 m

           =1.9306 x10-2J

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When the bob passes through the lower position, the energy iscompleted in the form of kinetic energy

      (1/2)mV2 =1.9306x10-2J

             V = 0.39299 m/s

                 = 0.39 m/s

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