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Oscillations Examples

Oscillations Examples. 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 of x , A = 5 cm = 0.05 m

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Oscillations Examples

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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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