A 2.75 kg particle moves as function of time as follows: x(t) = 5cos(1.25t+π/4); where distance is measured in meter and time in seconds. (a) What is the amplitude, frequency, angular frequency, and period of this motion? (b) What is the equation of the velocity of this particle? (c) What is the equation of the acceleration of this particle? (d) What is the spring constant? (e) What are the equations for the potential and kinetic energies of the particle? (f) What is the total energy?

Respuesta :

This question involves the concept of simple harmonic motion.

a) amplitude = 5m

frequency = 0.199 /s

angular frequency = 1.25 rad/s

period = 5.0277 s

b) Equation of velocity; v = -6.25sin(1.25t + (π/4))

c) equation of the acceleration; a = 7.8125 sin(1.25t + (π/4))

d) spring constant = 4.2969 N/m

e) Potential energy equation; U = 53.711cos²(1.25t + [tex]\frac{\pi}{4}[/tex])

kinetic energy equation; K = 53.711sin²(1.25t + (π/4))

f) total energy = 53.711 J

The general expression of the phase form of simple harmonic motion is;

x(t) = Acos(ωt + φ)

where;

A is amplitude

ω is angular frequency

t is time

φ is phase constant

We are given;

Mass of particle; m = 2.75 kg

Motion of particle; x(t) = 5cos(1.25t + (π/4))

  • a) Comparing this general expression to the one we are given in the question, we can see that;

A = 5 m

ω = 1.25 rad/s

φ = π/4

formula for frequency is;

f =  ω/2π

f = 1.25/2π

f = 0.199 /s

Period is given by the formula;

T = 1/f

T = 1/0.1989

T = 5.0277 s

  • b) From first derivative of the position equation, the velocity expressed with respect to time is given as;

v = −ωAsin(ωt + φ)

Plugging in the relevant values, we have;

v = -1.25 × 5 sin(1.25t + (π/4))

v = -6.25sin(1.25t + (π/4))

  • c) From second derivative of the position equation, the acceleration with respect to time is given as;

a = −ω²Acos(ωt + φ)

Plugging in the relevant values, we have;

a = [tex]1.25^{2}[/tex] × 5 sin(1.25t + (π/4))

a = 7.8125 sin(1.25t + (π/4))

  • d) Formula for the spring constant is;

k = ω²m

k = 1.25² × 2.75

k = 4.2969 N/m

  • e) The potential energy is given by the formula;

U = ½Kx²

U = ½(4.2969)[4cos(1.25t + π/4)]²

U = 53.711cos²(1.25t + [tex]\frac{\pi}{4}[/tex])

The kinetic energy is given by the formula;

K = ½m[tex]v^{2}[/tex]

K = ½ × 2.75 × [-6.25sin(1.25t + (π/4))]²

K = 53.711sin²(1.25t + (π/4))

  • F) The total energy is the sum of the potential and kinetic energy. Thus;

E =  53.711cos²(1.25t + [tex]\frac{\pi}{4}[/tex]) + 53.711sin²(1.25t + (π/4))

E = 53.711[cos²(1.25t + [tex]\frac{\pi}{4}[/tex]) + sin²(1.25t + (π/4))]

From trigonometry, we know that;

sin²x + cos²x = 1

Thus;

E = 53.711 J

Read more at; brainly.com/question/12722639