Question

A metal 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

Answer 1

Answer:

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