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A solid metal block with a mass of 4.30 kg is attached to a spring and is able to oscillate horizontally with negligible friction. The block 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 block oscillates along the x-axis, where x = 0 is the equilibrium position.) a) What is the spring constant (in N/m)?b) What is the frequency of the oscillations (in Hz)?c) What is the maximum speed of the block (in m/s)?d) At what position(s) (in m) on the x-axis does the maximum speed occur?

Respuesta :

Answer:

a) k = 120 N / m n b) f = 0.84 Hz , c)   v = 1,056 m / s , d)  x=0

Explanation:

a) To find the spring constant let's use Hooke's law

         F = - k x

         k = -F / x

         k = - 24.0 / (0.2)

         k = 120 N / m

b) the angular velocity in harmonic motion is

        w = √ k / m

        w = √ 120 / 4.30

        w = 5.28 rad / s

The angular velocity is related to the frequency

       w = 2π f

        f = w / 2π

        f = 5.28 / 2π

        f = 0.84 Hz

c) System speed is

        v = dx / dt = -A w sin (wt +φ)

The speed is maximum when sin (wt + φ) = ±1

       v = A w

       v = 0.200 5.28

       v = 1,056 m / s

d) in which position the velocity is maximum, as the velocity is a function of the sine and the position of the cosine when the sine is maximum the cosine is zero, so

       x = A cos 0 = 0

        x=0