Ametal cylinder with a mass of 4.20 kg is attached to a spring and is able to oscillate horizontally with negligible friction. the cylinder is pulled to a distance of 0.200 m from its equilibrium position held in place with a force of 24.0 n, and then released from rest. it then oscillates in simple harmonic motion. (the cylinder oscillates along the x-axis, where x0 is the equilibrium position.)

(a) what is the spring constant (in n/m)? n/m

(b) what is the frequency of the oscillations (in hz)? hz (c) what is the maximum speed of the cylinder (in m/s)?
m/s

(d) at what position(s) (in m) on the x-axis does the maximum speed occur?

(e) what is the maximum acceleration of the cylinder? (enter the magnitude in m/s2.)

(f) at what position(s) (in m) on the x-axis does the maximum acceleration occur?

(g) what is the total mechanical energy of the oscillating spring-cylinder system (in j)?
j
(h) what is the speed of the cylinder (in m/s) when its position is equal to one-third of the maximum displacement from equilibrium?
m/s
(i) what is the magnitude of the acceleration of the cylinder (in m/s2) when its position is equal to one-third of the maximum displacement from equilibrium?
m/s2

a) k = 120 N / m

, b)    f = 0.851 Hz

, c)  v = 1,069 m / s

, d)  x = 0

, e)  a = 5.71 m / s²

, f)   x = 0.200 m

, g)  Em = 2.4 J

, h) v = -1.01 m / s

Explanation:

a) Hooke's law is

         F = k x

         k = F / x

          k = 24.0 / 0.200

          k = 120 N / m

b) the angular velocity of the simple harmonic movement is

        w = √ k / m

        w = √ (120 / 4.2)

        w = 5,345 rad / s

Angular velocity and frequency are related.

       w = 2π f

        f = w / 2π

        f = 5.345 / 2π

        f = 0.851 Hz

c) the equation that describes the movement is

        x = A cos (wt + Ф)

As the body is released without initial velocity, Ф = 0

        x = 0.2 cos wt

Speed is

       v = dx / dt

       v = -A w sin wt

The speed is maximum for sin wt = ±1

       v = A w

       v = 0.200 5.345

       v = 1,069 m / s

d) when the function sin wt = -1 the function cos wt = 0, whereby the position for maximum speed is

       x = A cos wt = 0

       x = 0

e) the acceleration is

       a = d²x / dt² = dv / dt

       a = - Aw² cos wt

The acceleration is maximum when cos wt = ± 1

       a = A w²

        a = 0.2   5.345

        a = 5.71 m / s²

f) the position for this acceleration is

       x = A cos wt

       x = A

       x = 0.200 m

g) Mechanical energy is

        Em = ½ k A²

        Em = ½ 120 0.2²

       Em = 2.4 J

h) the position is

         x = 1/3 A

Let's calculate the time to reach this point

         x = A cos wt

        1/3 A = A cos 5.345t

         t = 1 / w cos⁻¹(1/3)

The angles are in radians

t = 1.23 / 5,345

t = 0.2301 s

Speed is

v = -A w sin wt

v = -0.2 5.345 sin (5.345 0.2301)

v = -1.01 m / s

i) acceleration

a = -A w² sin wt

a = - 0.2 5.345² cos (5.345 0.2301)

      a = -1.91 m / s²

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.

increase masses, decrease separation