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

Kinetic energy at 0.05 m is 0.037 J

Step-by-step 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(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=(1)/(2) kA^(2)\sin^(2)(\omega t)`

Substitute the suitable values in the above equation.

`E=(1)/(2) * 10 * 0.1^(2) \sin^(2)(2.24*0.47)`

E = 0.037 J