A 0.500-kg particle is attached to a spring (k = 50.0 N/m). At time t = 0, the p

Question

A 0.500-kg particle is attached to a spring (k = 50.0 N/m). At time t = 0, the particle has its
                    maximum speed of 20.0 m/s and is moving to the left.
                    (a) Determine the particle's equation of motion, specifying its position as a function of time.
                    (b) Where in the motion is the potential energy three times the kinetic energy?
                    (c) What time interval is required for the particle to move from x = 0 to x = 1.00 m?
                    (d) Find the length of a simple pendulum with the same period.

Explanation / Answer

w=sqrt(k/m)=sqrt(50/0.5)=10

Max speed=Aw=20 m/s

==> A=20/10=2 m

a) ==> equation:: y=2*sin(10*t)

b) 0.5*Kx^2=3*1/2*mv^2

==> 0.5*Kx^2 +0.5*mv^2=0.5*m*20^2

==> 4*0.5*Kx^2/3=0.5*0.5*20^2

==> x=sqrt(3*0.5*20^2/(50*4))= 1.73 m

c) 1=2*sin(10*t)

==> t= 0.52/10=0.052 sec

d) 2*pi/10=2*pi*sqrt(l/g)

==> l=g/100=0.098 m= 9.8 cm

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