Question

A spring of negligible mass stretches 3.00 cm from its relaxed
length when a force of 7.30 N is applied. A 0.520-kg particle rests
on a frictionless horizontal surface and is attached to the free
end of the spring. The particle is displaced from the origin to
*x* = 5.00 cm and released from rest at *t* = 0.
(Assume that the direction of the initial displacement is positive.
Use the exact values you enter to make later calculations.)

(f) Determine the displacement *x* of the particle from the
equilibrium position at *t* = 0.500 s.

cm

(g) Determine the velocity and acceleration of the particle when
*t* = 0.500 s. (Indicate the direction with the sign of your
answer.)

v = |
m/s |

a = |
m/s^{2} |

Answer #1

7.3 N force stretches 3 cm

so,

force = kx

7.3 = k * 0.03

k = 243.33 N/m

angular frequency = sqrt(k / m)

angular frequency = sqrt(243.33 / 0.52)

angular frequency = 21.63 rad/sec

equation of particle displacement

y(t) = A * cos(wt)

A = 5 cm

at time t = 0.5 sec

y(0.5) = 0.05 * cos(21.63 * 0.5)

**f) y(0.5) = -0.00897 m or -0.897 cm**

velocity(t) = -Aw * sin(wt)

velocity(0.5) = -0.05 * 21.63 * sin(21.63 * 0.5)

**g) velocity(0.5) = 1.0639 m/s**

acceleration(t) = Aw^2 * cos(wt)

acceleration(0.5) = 0.05 * 21.63^2 * cos(21.63 * 0.5)

**g) acceleration(0.5) = -4.201 m/s^2**

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