A pendulum of mass 2 kg and length 2 m is displaced 10 cm from its equilibrium p

Question

A pendulum of mass 2 kg and length 2 m is displaced 10 cm from its equilibrium position and released, (a) What is the equation giving the horizonatal force F(x) required to displace the mass a horizontal distance x for small values of x? (b) Is this equation valid for x = 10 cm? Why or why not? (c) What is the effective spring constant of the pendulum? (c) What is the period of the pendulum? (d) What is the equation of motion for this pendulum? (e) What is the kinetic energy of the pendulum when it is 5 cm from the equilibrium position?

Explanation / Answer

FORCE F=-kx

in a simple harmonic motion force is proportional to the negative dispalcement of the pendulum from the equlibrium position

F=mg=kx

k=2*10/0.1=200 N/m

time period t=2*pisqrt(m/k)

t=0.628 sec

equation of motion:f=-kx

d^2x/dt^2+kx=0

kinetic energy=1/2*mv^2

v=Aw

here A=10/2=5 CM=0.05 m

K=1/2*200*0.05^2 (1/2KA^2)

K=0.25 J

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