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.
[tex]\omega = \sqrt\frac{10}{2}[/tex]
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.
[tex]0.05 = 0.1\cos(2.24t)[/tex]
t = 0.47 s
Kinetic energy at x = 0.05 is determine by the relation:
[tex]E=\frac{1}{2} kA^{2}\sin^{2}(\omega t)[/tex]
Substitute the suitable values in the above equation.
[tex]E=\frac{1}{2} \times 10 \times 0.1^{2} \sin^{2}(2.24\times0.47)[/tex]
E = 0.037 J
