Question

A 2 kg mass connected to a spring with spring constant k = 10 N/m oscillates in simple harmonic motion with an amplitude of A = 0.1 m. What is the kinetic energy of the mass when its position is at x = 0.05 m? g

Answer 1

Answer:

Kinetic energy at 0.05 m is 0.037 J

Explanation:

Given:

Mass, m = 2 kg

Spring constant, k = 10 N/m

Amplitude, A = 0.1 m

Angular frequency, ω = √k/m

Substitute the suitable values in the above equation.

$\omega = \sqrt\frac{10}{2}$

ω = 2.24 s⁻¹

Simple harmonic equation is represent by the equation:

x = A cos ωt

Substitute 0.05 m for x, 0.1 m for A and 2.24 s⁻¹ for ω in the above equation.

$0.05 = 0.1\cos(2.24t)$

t = 0.47 s

Kinetic energy at x = 0.05 is determine by the relation:

$E=\frac{1}{2} kA^{2}\sin^{2}(\omega t)$

Substitute the suitable values in the above equation.

$E=\frac{1}{2} \times 10 \times 0.1^{2} \sin^{2}(2.24\times0.47)$

E = 0.037 J