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Harmonic Motion

Harmonic Motion. Chapter 13. Simple Harmonic Motion. A single sequence of moves that constitutes the repeated unit in a periodic motion is called a cycle The time it takes for a system to complete a cycle is a period (T). Simple Harmonic Motion.

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Harmonic Motion

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  1. Harmonic Motion Chapter 13

  2. Simple Harmonic Motion • A single sequence of moves that constitutes the repeated unit in a periodic motion is called a cycle • The time it takes for a system to complete a cycle isa period (T)

  3. Simple Harmonic Motion • The period is the number of units of time per cycle; the reciprocal of that—the number of cycles per unit of time—is known as the frequency(f). • The SI unit of frequency is the hertz (Hz), where 1 Hz = 1 cycle/s = 1 s-1 • Amplitude (A) is the maximum displacement of an object in SHM

  4. Simple Harmonic Motion • One complete orbit (one cycle) object sweeps through 2prad • f - number of cycles per second • Number of radians it movesthrough per second is 2pf - that's angular speed (w) • w - angular frequency • Sinusoidal motion (harmonic) with a single frequency - known as simple harmonic motion (SHM)

  5. Displacement in SHM

  6. Velocity in SHM vmax = Aw

  7. Acceleration in SHM The acceleration of a simple harmonic oscillator is proportional to its displacement

  8. Example 1 • A spot of light on the screen of a computer is oscillating to and fro along a horizontal straight line in SHM with a frequency of 1.5 Hz. The total length of the line traversed is 20 cm, and the spot begins the process at the far right. Determine • (a) its angular frequency, • (b) its period, • (c) the magnitude of its maximum velocity, and • (d) the magnitude of its maximum acceleration, • (e) Write an expression for x and find the location of the spot at t = 0.40 s.

  9. Example 1 • Given: f = 1.5 Hz and A = 10 cm • Find: (a) w, (b) T, (c) |vx(max)|, (d) |ax(max)|, (e) x in general, and x at t = 0.40s From Eq. (10.12) (a) w = 2pf = 2p(1.5 Hz) = 9.4 rad/s = 3.0prad/s (b) T= 1/f= 1/1.5 Hz = 0.67s (c) |vx(max)| = vmax = Aw= 2pfA vmax = 2p(1.5 Hz)(0.10 m) = 0.94 m/s

  10. Example 1 (d) From Eq. (10.17), |ax(max)| = Aw2= A(2pf)2= (0.10 m)(2p 1.5 Hz)2 = 8.9 m/s2 (e) x = A coswt = (0.10 m) cos (9.4 rad/s)t At t = 0.40 s x = (0.10m) cos (3.77 rad) = (0.10 m)(-0.81) = -8.1 cm

  11. Problem • A point at the end of a spoon whose handle is clenched between someone’s teeth vibrates in SHM at 50Hz with an amplitude of 0.50cm. Determine its acceleration at the extremes of each swing.

  12. Equilibrium • The state in which an elastic or oscillating system most wants to be in if undisturbed by outside forces.

  13. Elastic Restoring Force • When a system oscillates naturally it moves against a restoring force that returns it to its undisturbed equilibrium condition • A "lossless" single-frequency ideal vibrator is known as a simple harmonic oscillator.

  14. An Oscillating Spring • If a spring with a mass attached to it is slightly distorted, it will oscillate in a way very closely resembling SHM. • Force exerted by an elastically stretched spring is the elastic restoring force F, = -ks. • Resulting acceleration ax = -(k/m)x • F is linear in x; a is linear in x - hallmark of SHM

  15. Frequency and Period Simple harmonic oscillator Shown every ¼ cycles for 2 cycles Relationship between x, vx, t, and T

  16. Hooke’s Law • Beyond being elastic, many materials deform in proportion to the load they support - Hooke's Law

  17. Hooke’s Law • The spring constant or elastic constant k - a measure of the stiffness of the object being deformed • k hasunits of N/m

  18. Hooke’s Law • k hasunits of N/m

  19. Frequency and Period • w0 - the natural angular frequency, the specific frequency at which a physical system oscillates all by itself once set in motion natural angular frequency • and since w0 = 2pf0 natural linear frequency • Since T= 1/f0 Period

  20. Resonance vs. Damping • If the frequency of the disturbing force equals the natural frequency of the system, the amplitude of the oscillation will increase—RESONANCE • If the frequency of the periodic force does NOT equal the natural frequency of the system, the amplitude of the oscillation will decrease--DAMPING

  21. Example 2 • A 2.0-kg bag of candy is hung on a vertical, helical, steel spring that elongates 50.0 cm under the load, suspending the bag 1.00 m above the head of an expectant youngster. The candy is pulled down an additional 25.0 cm and released. How long will it take for the bag to return to a height of 1.00 m above the child?

  22. Example 2 • Given: m = 2.0 kg, ∆L = 50.0 cm, and A = 25.0 cm. Find: t when y = 0. • To find T, we must first find k • Initially the load (F = Fw) stretches the spring ¼T = 0.35 s

  23. The Pendulum • The period of a pendulum is independent of the mass and is determined by the square root of its length

  24. Example 3 • How long should a pendulum be if it is to have a period of 1.00 s at a place on Earth where the acceleration due to gravity is 9.81 m/s2? • Given: T = 1.00 s and g = 9.81 m/s2. Find: L • From T2= 4π2 (L/g); hence

  25. Problem 2 • What would the length of a pendulum need to be on Jupiter in order to keep the same time as a clock on Earth? gJupiter = 25.95m/s2

  26. When “f” is known When “f” is NOT known vmax = Aw

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