A pendulum of mass 2 kg at length 2 m is displaced 10 cm from its equilibrium po

Question

A pendulum of mass 2 kg at length 2 m is displaced 10 cm from its equilibrium position and released.

(A) What is the equation giving the horizontal force F(x) required to display the mass a horizontal distance x for small values of x?

(B) Is the equation valid for x=10cm? why ?

(C) What is the effective spring constant of the pendulum?

(D) What is the period of the pendulum?

(E) What is the equation of the motion for this pendulum?

(F) 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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