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Simple Harmonic Motion: Mass-Spring System Analysis and Energy Transfer

This example explores a mass-spring system demonstrating simple harmonic motion (SHM) with a mass of 0.2 kg and a spring constant of 5 N/m. The initial displacement is 0.05 m. Key calculations include amplitude, angular frequency, frequency, period, maximum speed, and maximum acceleration. The dynamics of motion as functions of time are derived, juxtaposed with energy conservation principles in SHM. By examining modifications to initial conditions, such as adding initial velocity, we explore how these changes affect the system's behavior and energy distribution.

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Simple Harmonic Motion: Mass-Spring System Analysis and Energy Transfer

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  1. Oscillations Examples

  2. Example 15.1 • Mass-Spring System: m = 200 g = 0.2 kg, k = 5 N/m. • Moved x = 5 cm = 0.05 m & let go on frictionless surface. Simple harmonic motion. Find: A. Amplitude: A= maximum value ofx, A = 5 cm = 0.05 m B. Angular frequency ω: so ω = 5 rad/s Frequency f: so f = ω/(2π) = 0.8 Hz (cycles/second) Period T: From above, T = (1/f) = 1.26 s C.Max. speed vmax & max. acceleration amax: so, vmax= 0.25 m/s amax= 1.25 m/s2

  3. Example 15.1, continued • Mass-Spring System: m = 200 g = 0.2 kg, k = 5 N/m. Found: ω = 5 rad/s, A = 0.05 m Find: D. Position, velocity, & acceleration as functions of time: x(t) = Acos(wt + f), v(t) = - wAsin(wt + f), a(t) = - w2Acos(wt + f) To find f note when t = 0, x = x(0) = A = Acos f so, f = 0 So, x(t) = Acos(wt) = (0.05 m)cos(5t) v(t) = - wAsin(wt) = - (0.25 m/s)sin(5t) a(t) = - w2Acos(wt) = - (1.25 m/s2)cos(5t)

  4. Modification of Example 15.1 • Mass-Spring System: m = 200 g = 0.2 kg, k = 5 N/m. Instead of pullingmout a distance xi = 0.05 m & releasing it with zero initial velocity, pull it that distance & push it at an initial velocityvi = -0.1 m/s. Which parts of problem change? Still have: x(t) = Acos(wt + f), v(t) = - wAsin(wt + f), a(t) = - w2Acos(wt + f) Angular frequency ω: is stillω = 5 rad/s Frequency f is still= ω/(2π) = 0.8 Hz (cycles/second) Period T is still= (1/f) = 1.26 s

  5. Modification Continued • Mass-Spring System: m = 200 g = 0.2 kg, k = 5 N/m. ω = 5 rad/s, xi = 0.05 m, vi = - 0.1 m/s x(t) = Acos(wt + f), v(t) = - wAsin(wt + f), a(t) = - w2Acos(wt + f) Everything else changes! Find f & A: when t = 0, x = x(0) = xi = 0.05 m = Acosf (1) and v = v(0) = vi = - 0.1 m/s = - ωAsinf (2) (1) & (2): 2 equations, 2 unknowns. Algebra gives: tanf = - vi/(wxi) = 0.4, so f = 0.127π radians A = xi/cosf = 0.054 m so, vmax= ωA = 0.271 m/s, amax= ω2A = 1.36 m/s2 So, x(t) = Acos(wt + f) = (0.054 m)cos(5t + 0.127π) v(t) = wAsin(wt + f) = - (0.271 m/s)sin(5t + 0.127π) a(t) = - w2Acos(wt) = - (1.36 m/s2)cos(5t + 0.127π)

  6. Energy of the SHM Oscillator • Assume a spring-mass system is moving on a frictionless surface • This tells us the total energy is constant • The kinetic energy can be found by • K = ½ mv 2 = ½ mw2A2 sin2 (wt + f) • The elastic potential energy can be found by • U = ½ kx 2 = ½ kA2 cos2 (wt + f) • The total energy is E =K + U = ½ kA 2

  7. The total mechanical energy is constant • The total mechanical energy is proportional to the square of the amplitude • Energy is continuously being transferred between potential energy stored in the spring and the kinetic energy of the block • Use the active figure to investigate the relationship between the motion and the energy

  8. As the motion continues, the exchange of energy also continues • Energy can be used to find the velocity

  9. Energy in SHM, summary

  10. Importance of Simple Harmonic Oscillators • Simple harmonic oscillators are good models of a wide variety of physical phenomena • Molecular example • If the atoms in the molecule do not move too far, the forces between them can be modeled as if there were springs between the atoms • The potential energy acts similar to that of the SHM oscillator

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