A 1.90-kg object is attached to a spring and placed on frictionless, horizontal

Question

A 1.90-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 21.0 N is required to hold the object at rest when it is pulled 0.200 m from its equilibrium position (the origin of the x axis). The object is now released from rest from this stretched position, and it subsequently undergoes simple harmonic oscillations. (a) Find the force constant of the spring.
N/m

(b) Find the frequency of the oscillations.
Hz

(c) Find the maximum speed of the object.
m/s

(d) Where does this maximum speed occur?
x = ± m

(e) Find the maximum acceleration of the object.
m/s2

(f) Where does the maximum acceleration occur?
x = ± m

(g) Find the total energy of the oscillating system.
J

(h) Find the speed of the object when its position is equal to one-third of the maximum value.
m/s

(i) Find the magnitude of the acceleration of the object when its position is equal to one-third of the maximum value.
m/s2

Explanation / Answer

(a) k = F / x = 21.0N / 0.200m = 105 N/m

(b) = (k/m) = (105N/m / 1.90kg) = 7.434 rad/s

(c) max speed = A* = 0.200m * 7.434rad/s = 1.487 m/s

(d) Max speed a x = 0.00 m

(e) max acceleration = A² = 0.200m * (7.434rad/s)² = 11.053 m/s²

(f) Max acceleration at x = ±0.200 m

(g) TME = max PE = ½kA² = ½ * 105N/m * (0.200m/s)² = 2.10 J

H) Use 1/2*k*x^2 + 1/2*m*v^2 = 1/2*m*v_max^2 = 1/2*k*A^2 and just solve for v. Then use x = A/3.

I) For this part, use the SHM equations
v(t) = w*A*cos(w*t + phi)
a(t) = -w^2*A*sin(w*t + phi)

Assume phi = 0. Using the v(t) you just found in part H, solve for the time. Then plug the time into a(t) equation to get the acceleration.

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