The length of a simple pendulum is 0.84 m and the mass of the particle (the...

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The length of a simple pendulum is 0.84 m and the mass of theparticle (the \"bob\") at the end of the cable is 0.23 kg. Thependulum is pulled away from its equilibrium position by an angleof 9.05° and released from rest. Assume that friction can beneglected and that the resulting oscillatory motion is simpleharmonic motion.

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from above data that

The Time period of the pendulum

       T = 2π √ [ L/g ]

          = 2π √[0.84 /9.8 ]

          =1.84 s

The angular frequency is

        ω= 2π / T

            =2 π / 1.84

           = 3.41 rad/ s

The height through which the bob is lifted up is

        h = L ( 1 - cosθ )

           =0.84 ( 1 - cos 9.05)

           =0.0104 m

The total mechanical energy is

        E = m g h

           =0.23(9.8)0.0104 m

           =2.34 x10-2J

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

  0.5*m*v²;

v = √[2*KE/m] = √[2*0.0234/0.23] = 0.461 m/s


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