Ejercicio Exercise 1 - Simple harmonic motion · Física II URV 2024

Tema: Oscillations: simple harmonic, damped and forced · Del examen Extraordinary Exam - Physics (2024)

Enunciado

Exercise 1 (10 points). Mass-spring system and damped oscillations

A mass of 200 g200\text{ g} is suspended from a system of two parallel springs as illustrated below. The elastic constants of the two springs are k1=0.9 N/mk_1 = 0.9\text{ N/m} and k2=1.1 N/mk_2 = 1.1\text{ N/m}, respectively.

At time t=0t = 0, the mass is released from rest at a position −2 cm-2\text{ cm} away from equilibrium. Assuming negligible air resistance:

a.

Determine the equivalent elastic constant of the assembly, as well as the pulsation (angular frequency) and the oscillation period. [2 points]

b.

Write the equation of motion x(t)x(t) in SI units (express xx in meters). [2 points]

c.

Compute the total mechanical energy of the system at any given moment. How is this energy divided between kinetic and potential forms when the mass reaches +1 cm+1\text{ cm} from the equilibrium position? [2 points]

Enunciado: Ejercicio Exercise 1 - Simple harmonic motion · Física II URV 2024

Solución

Según los criterios oficiales de corrección

a. Equivalent elastic constant of the assembly keqk_{\text{eq}}

Solución: 2.02.0 N/m

a. Pulsation (angular frequency) ω0\omega_0

Solución: 10\sqrt{10} rad/s / 3.1623.162 rad/s / 3.1623.162 s⁻¹

a. Oscillation period T0T_0

Solución: 1.9871.987 s

b. Write the equation of motion x(t)x(t)

Solución: −0.02cos⁡(10t)-0.02\cos(\sqrt{10} t) / 0.02cos⁡(10t+π)0.02\cos(\sqrt{10} t + \pi) / −0.02cos⁡(3.162t)-0.02\cos(3.162 t)

c. Total mechanical energy EE

Solución: 4.0⋅10−44.0\cdot 10^{-4} J / 0.40.4 mJ

c. Elastic potential energy EpE_p at x=+1 cmx = +1\text{ cm}

Solución: 1.0⋅10−41.0\cdot 10^{-4} J / 0.10.1 mJ

c. Kinetic energy EkE_k at x=+1 cmx = +1\text{ cm}

Solución: 3.0⋅10−43.0\cdot 10^{-4} J / 0.30.3 mJ

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