# Harmonic Motion Chapter 13 Simple Harmonic Motion A

• Slides: 26

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 �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

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

Displacement in SHM

Velocity in SHM vmax = Aw

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

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.

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. 40 s From Eq. (10. 12) (a) w = 2 pf = 2 p(1. 5 Hz) = 9. 4 rad/s = 3. 0 p rad/s (b) T= 1/f= 1/1. 5 Hz = 0. 67 s (c) |vx(max)| = vmax = Aw = 2 pf. A vmax = 2 p(1. 5 Hz)(0. 10 m) = 0. 94 m/s

Example 1 (d) From Eq. (10. 17), |ax(max)| = Aw 2 = A(2 pf)2= (0. 10 m)(2 p 1. 5 Hz)2 = 8. 9 m/s 2 (e) x = A cos wt = (0. 10 m) cos (9. 4 rad/s)t At t = 0. 40 s x = (0. 10 m) cos (3. 77 rad) = (0. 10 m)(-0. 81) = -8. 1 cm

Problem �A point at the end of a spoon whose handle is clenched between someone’s teeth vibrates in SHM at 50 Hz with an amplitude of 0. 50 cm. Determine its acceleration at the extremes of each swing.

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

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.

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 n F is linear in x; a is linear in x - hallmark of SHM

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

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

Hooke’s Law n n k has units of N/m The spring constant or elastic constant k - a measure of the stiffness of the object being deformed

Hooke’s Law n k has units of N/m

Frequency and Period �w 0 - 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 w 0 = 2 pf 0 natural linear frequency �Since T= 1/f 0 Period

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

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?

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

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

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/s 2? �Given: T = 1. 00 s and g = 9. 81 m/s 2. Find: L �From T 2 = 4π2 (L/g); hence

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? g. Jupiter = 25. 95 m/s 2

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