Question

The length of a simple pendulum is 0.84 m and the mass of the particle (the "bob") at the end of the cable is 0.23 kg. The pendulum is pulled away from its equilibrium position by an angle of 9.05° and released from rest. Assume that friction can be neglected and that the resulting oscillatory motion is simple harmonic motion.

Answer 1

ans)

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