Download
1 / 49
490 likes | 499 Views
Learn about oscillations, SHM dynamics, key equations, and examples in this comprehensive guide. Discover characteristics, graphs, and applications of oscillating systems.
E N D
Chapter 15 Oscillations
k m k m k m Oscillations Let’s assume we have a horizontal spring and mass system composed of a spring with the elastic (spring) constant k, and a mass m. • If we stretch the spring with the mass on the end and let it go, the mass will continuously move back and forth (if there is no friction). • The back and forth or up and down motion that repeats itself on the same path at equal intervals of time is called oscillation (vibration).
Characteristics of Oscillations Horizontal Spring x -stretch- displacement from the equilibrium (unstretched) position. Maximum displacement- amplitudeA • The motion will repeat itself after an interval of time T called period. • T is measured in seconds s. The number of oscillations per unit of time is called frequency f. It is measured in Hz or s-1.
If n oscillations are happening in the time t, then: • T = t/n, and f = n/t. • Therefore T= 1/f, f = 1/T, or fT =1 Example T, f • A mass oscillates 20 times in 5s. What is the period and the frequency of the oscillation? T = 5s/20 = 0.25 s f = 20/5s = 4 Hz Check Tf = 1, 4 Hz x .25 s = 1
F = -kx a k m x SHM Dynamics At any given instant the force F = -kxcalled restoring force acts on the mass. It is proportional to the displacement, opposes motion and points towards the equilibrium position. An oscillation caused by a restoring force is called Simple Harmonic Motion –SHM.
F = -kx a k m x SHM Dynamics • At any given instant we know thatF = mamust be true. • But in this case F = -kx and ma = • So: -kx = ma = a differential equation for x(t)!
Solution for the differential equation: • Where: • A = amplitude: max displacement from equilibrium position • = angular frequency • t + δ = phase • δ = phase constant (phase @ t =0)
Comparison b/w two oscillations: • Lets assume we have two oscillations: and If δ = 2nπ, n= …-2,-1,0,1,2,… than: x1=x2, and the systems are in phase. If δ = (2n+1)π, n= …-2,-1,0,1,2,… than: x1=-x2, and the systems are (180o)out of phase.
Citicorp building in New York. Spring and 400 ton mass system oscillates out of phase with building reducing the swaying during high winds (same natural frequency).
The derivative of x gives the velocity v: Differentiating the velocity with respect to the time we get the acceleration: 14-6 Since: , we get a relationship b/w the position and acceleration:
Comparing a=-(k/m)x with a = -ω2x we get: A and δ can be determined from the initial conditions xo and vo. Setting t=0 in: gives: Square and add: Also:
The object is at the same location after one period T: x(t) = x(t+T) Since cos repeats in value after 2π we must have: Solving for f: or: Or the period:
Astronaut measures his mass by sitting in a seat attached to a spring and oscillating back and forth. The total mass of astronaut + seat is related to the frequency of oscillation by the equation:
Bubbles from a propeller rotating in water produce a sinusoidal pattern.
An object oscillates with ω = 8 rad/s. At t = 0, the object is at xo = 4 cm, with the initial velocity of – 25 cm/s. a) Find the amplitude and the phase constant for this motion. Use with t =0: Rearranging: Divide
Graph of U vs. x. The blue line is the total energy. The kinetic energy is the vertical distance K=Etotal-U
Object on Vertical Spring: Unstretched: Object oscillating: Change variable: y=yo+y’ Standard SHM equation with solution:
Example 1 f A small mass m1 rests on but is not attached to a large mass M2 that slides on its base without friction. The maximum frictional force between m1 and M2 is f. A spring of spring constant k is attached to the large mass M2 and to the wall as shown. a. Determine the maximum horizontal acceleration that M2 may have without causing m1 to slip. amax = f/m1 b. Determine the maximum amplitude A for simple harmonic motion of the two masses if they are to move together, i.e., m1 must not slip on M2. In m = m1 + M2 amax = ω2A
O x Example 1 Continuation f -kx c. The two‑mass combination is pulled to the right the maximum amplitude A found in part (b) and released. Describe the frictional force on the small mass m1 during the first half cycle of oscillation. c) f = m1a, follows acceleration, decrease for T/4, than increase for T/4 d. The two‑mass combination is now pulled to the right a distance of A' greater than A and released. i. Determine the acceleration of m1 at the instant the masses are released. amax = ω2A’ ii. Determine the acceleration of M2 at the instant the masses are released. f-kA’ = M2a2
Example 2 A block rests on a spring and oscillates vertically with 4Hz, and amplitude 7 cm. A tiny bead is placed on the top of the oscillating block as it reaches its lowest point. Assume that the bead’s mass is negligible. At what distance from the block’s equilibrium position will the bead lose contact with the block? N mg ay -N + mg = may Lose contact at N=0 ay = g
A uniform rod of mass M and length L is free to rotate about a horizontal axis perpendicular to the rod and through one end of the rod. a) Find the period of oscillation for small angular displacements. D = L/2
A uniform rod of mass M and length L is free to rotate about a horizontal axis perpendicular to the rod. b) Find the period of oscillation if the rotation axis is a distance x from the CM. D = x
