Suppose you have a particle that starts at x = 1 m and oscillates around an equi

Question

Suppose you have a particle that starts at x = 1 m and oscillates around an equilibrium position of x = 0 with a period of 1 sec. What is the position as a function of t? For the example above, suppose the particle's mass is 1 kg. What is the kinetic energy as a function of time? potential energy? total energy? A spring that hangs on its initial equilibrium position gets extended by 0.1 m after a 1 kg mass is hung on its end. What is the period of oscillation for this system if the mass is pulled down a little?

Explanation / Answer

a) x(t)=A*cos(w*t) =1*cos(2*PI/1*t) =cos(6.28*t)

b) m=1kg,

v(t)= dx/dt =-6.28*sin(6.28t)

K.E= 1/2*m*v^2 =1/2*1*(-6.28*sin(6.28*t))^2=19.7192*sin(6.28*t)^2

P.E= 1/2*k*x^2=1/2*k*(cos(6.28*t))^2

T.E=1/2*k*(cos(6.28*t))^2+19.7192*sin(6.28*t)^2

C)A=0.1m, m=1kg

m*g =k*x

=>k=1*9.81/0.1 =98.1N/m

T=s2*PI*sqrt(m/k)=2*PI*sqrt(1/98.1)=0.63437398492s

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