Question

A 40.1 g object is attached to a horizontal spring with a spring constant of 11.9 N/m and released from rest with an amplitude of 24.7 cm. What is the velocity of the object when it is halfway to the equilibrium position if the surface is frictionless?

Answer 1

Answer: v = 3.684 m/s

Explanation: The angular frequency (ω) of a loaded spring is given as

ω = √k/m

Where ω = angular frequency, k =spring constant = 11.9 N/m, m = mass of object = 40.1 g = 0.0401 kg.

The velocity of a simple harmonic motion is defied as

v = ω√A² - x²

Where A = Amplitude = 24.7cm = 0.247m and x = displacement.

For our question, we where asked to find velocity at half way, at half way, x = A/2

Hence at half way, x = 0.247/2 = 0.1235 m.

We need to get the value of angular frequency first.

ω = √(11.9/0.0401)

ω = √296.758

ω = 17.22 rad/s.

Then the velocity is

v = 17.22 √0.247² - 0.1235²

v = 17.22 √0.061009 - 0.01525225

v = 17.22 √0.04575675

v = 17.22 × 0.2139

v = 3.684 m/s