the position of a 3kg object is given by x(t) = 10cm cos((2pi/3)t - (pi/2)) wher

Question

the position of a 3kg object is given by x(t) = 10cm cos((2pi/3)t - (pi/2)) where t is the time in seconds. (a) Calculate the total energy. (b) Calculate the time when the object first passes through it's equilibrium position. (C) calculate the time when the object first has a velocity of zero. (D) calculate the objects position and speed when the potential energy and kinetic energy are equal the position of a 3kg object is given by x(t) = 10cm cos((2pi/3)t - (pi/2)) where t is the time in seconds. (a) Calculate the total energy. (b) Calculate the time when the object first passes through it's equilibrium position. (C) calculate the time when the object first has a velocity of zero. (D) calculate the objects position and speed when the potential energy and kinetic energy are equal the position of a 3kg object is given by x(t) = 10cm cos((2pi/3)t - (pi/2)) where t is the time in seconds. (a) Calculate the total energy. (b) Calculate the time when the object first passes through it's equilibrium position. (C) calculate the time when the object first has a velocity of zero. (D) calculate the objects position and speed when the potential energy and kinetic energy are equal

Explanation / Answer

w = sqrt(k/m)

x(t) = xocos(wt - x )

compare to this with given equation

w = 2pi/3

m = 3kg

k = w^2*m

k = 13.146

part a ) Total energy = 1/2*k*A^2

A = 10 cm = 0.1 m

Etotal = 1/2 * 13.146 * 0.1 * 0.1

Etotal = 0.06573 J

part b ) at equalibrium

x = 0

cos((2pi/3)t - (pi/2)) = 0

(2pi/3)t - (pi/2) = (n+1/2)*pi

(2pi/3)t = (n+1/2)*pi + (pi/2)

t = ((n+1/2)*pi + (pi/2))/(2pi/3)

here n = 0

t = 1.5 s

part c )

x'(t) = -10cm *(2pi/3) sin((2pi/3)t - (pi/2))

sin((2pi/3)t - (pi/2)) = 0

sin((2pi/3)t - (pi/2)) = npi

n = 0

2pi/3 t = pi + pi/2

t = 2.25 s

part d )

1/2*k*x^2 = 1/2*mv^2

by solving above v and x equation

t = 8.79 s

x = 9.57 cm = 0.957 m

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