Question

A 0.20 kg object, attached to a spring with spring constant k = 10 N/m, is moving on a horizontal frictionless surface in simple harmonic motion of amplitude of 0.080 m. What is its speed at the instant when its displacement is 0.040 m?

Answer 1

V = 4.90m/s

Step-by-step explanation:

Given m = 0.20kg, k = 10N/m, A = 0.8m

x = 0.40m

The velocity of a body at a given displacement x having known the amplitude, mass and spring constant is given by the equation below.

V = √(k/m) × √(A²-x²)

So substituting the given values below, we have that

V = √(10/0.2) × √(0.8²-0.4²)

V = √(50) × √(0.48)

V = 4.90m/s.

Answer 2

The speed is 49 cm/s

Explanation:

Given;

mass of the object, m = 0.20 kg

spring constant, k = 10 N/m

amplitude, A = 0.080 m

Total energy of the spring = ¹/₂kA² = ¹/₂ x 10 x (0.08)²

Total energy of the spring = 0.032 J

When its displacement is 0.040 m = half of Amplitude

Applying conservation of energy

Total energy of the spring = PE + KE

0.032 = ¹/₂kA² + ¹/₂mv²

0.032 = ¹/₂ x 10 x (0.04)² + ¹/₂ x 0.2 x v²

0.032 = 0.008 + 0.1v²

0.1v² = 0.032 - 0.008

0.1v² = 0.024

v² = 0.24

v = √0.24

v = 0.49 m/s = 49 cm/s

Therefore, the speed is 49 cm/s