Question #161890

The period of a certain pendulum is 2.0 seconds and the mass of the pendulum Bob is 50g. The Bob is pulled aside through a horizontal distance of 8cm and then released. Find the displacement and kinetic energy of the Bob 0.7seconds after release


1
Expert's answer
2021-02-10T07:16:57-0500

Answer

Angular frequency

w=2πT=3.14w=\frac{2\pi }{T}=3.14 rad/sec

t=0.7sec

x0=0.08m

Amplitude is given

X=x0sinwtX=x_0sin wt

X=0.08×sin(3.14×0.7)X=0.08\times sin(3.14\times0.7)

X=0.064m

Now velocity

V=X×w=0.2V=X\times w=0.2 m/s

So kinetic energy

K=mV22=4×105K=\frac{mV^2}{2}=4\times10^{-5} J


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