* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download SIMPLE HARMONIC MOTION (1)
N-body problem wikipedia , lookup
Modified Newtonian dynamics wikipedia , lookup
Hooke's law wikipedia , lookup
Classical mechanics wikipedia , lookup
Fictitious force wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Statistical mechanics wikipedia , lookup
Brownian motion wikipedia , lookup
Electromagnetic mass wikipedia , lookup
Jerk (physics) wikipedia , lookup
Center of mass wikipedia , lookup
Relativistic mechanics wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Work (physics) wikipedia , lookup
Thermodynamic system wikipedia , lookup
Mass versus weight wikipedia , lookup
Hunting oscillation wikipedia , lookup
Equations of motion wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Classical central-force problem wikipedia , lookup
SIMPLE HARMONIC MOTION (1) • • • • • • Consider an air track with a cart of mass m attached to a spring of force constant k When the spring is at equilibrium length (neither stretched or compressed), the cart is at position x = 0 If the cart is displaced from equilibrium by a distance x, the spring exerts a restoring force given by F = -kx The spring exerts a restoring force whose magnitude is proportional to the distance it is displaced from equilibrium The force exerted by a spring is opposite in direction to its displacement from equilibrium; this accounts for the minus sign in F = -kx A restoring force is one that always points toward the equilibrium position 5. Oscillations About Equilibrium 1 SIMPLE HARMONIC MOTION (2) • • • • If the cart is released from rest at the location x = A, the spring exerts a force on the cart to the left, causing the cart to accelerate toward the equilibrium position When the cart reaches x = 0, the net force acting on it is zero, but its speed is not zero so it continues to move to the left As the cart compresses the spring, it experiences a force to the right, causing it to decelerate and come to rest at x = -A The spring continues to exert a force to the right; thus the cart immediately moves to the right until it comes to rest again at x = A, completing one oscillation in time T 5. Oscillations About Equilibrium 2 SIMPLE HARMONIC MOTION (3) • • • • • • • • • If a pen is attached to the cart, it can trace its motion on a sheet of paper moving with constant speed On this chart, we obtain a record of the cart’s motion as a function of time The motion of the cart looks like a sine or cosine function The position of the cart as a function of time can be represented by a sine or cosine function The position of the mass oscillates between x = +A and x = -A, where A is the amplitude of the motion and is the extreme displacement in each direction The amplitude is one half the total range of motion, which repeats with a period T The position of the cart is the same at time t + T as it is at time t Thus the mathematical description of position versus time in simple harmonic motion is x = Acos(2πt/T) (ensure calculator set to radians) 5. Oscillations About Equilibrium 3 SIMPLE HARMONIC MOTION: EXAMPLE • An air track cart attached to a spring completes one oscillation every 2.4s. At t = 0 the cart is released from rest at a distance of 0.1m from its equilibrium position. What is the position of the cart at 0.3s, 0.6s, 2.7s and 3.0s? 5. Oscillations About Equilibrium 4 CONNECTIONS BETWEEN UNIFORM CIRCULAR MOTION AND SIMPLE HARMONIC MOTION • • • • • A peg is placed at the rim of a turntable that rotates with constant angular velocity Viewed from above, the peg exhibits uniform circular motion Viewed from the side it appears to move back and forth in a straight line, as we can see by shining a light to cast a shadow of the peg onto a back screen The shadow of the peg moves with simple harmonic motion Thus there is a connection between uniform circular motion and simple harmonic motion 5. Oscillations About Equilibrium 5 SIMPLE HARMONIC MOTION: POSITION AS FUNCTION OF TIME • • • • • • Recall that angular position is simply θ = ωt Imagine drawing a radius vector of length A to the position of the peg When we project the shadow of the peg onto the back screen, the shadow is at a location x = Acosθ, which is the x component of the radius vector Thus for an object undergoing a simple harmonic motion, its position as a function of time is given by x = Acosθ = Acos(ωt) = Acos(2πt/T) When dealing with simple harmonic motion, ω is called the angular frequency ω = 2πf = 2π/T rad/s 5. Oscillations About Equilibrium 6 VELOCITY IN SIMPLE HARMONIC MOTION • • • • • • • • • Recall that the velocity of an object in uniform circular motion of radius r is v = rω The velocity is tangential to the object’s circular path, as indicated above When the peg is at the angular position θ, the velocity vector makes an angle θ with the vertical Thus the x component of the velocity is –vsinθ Thus vx = -vsinθ = -rωsinθ Since we are only concerned with movement in the x direction, that r = A and θ = ωt, the velocity in simple harmonic motion is v = -Aωsin(ωt) m/s Plotting x and v for SHM shows that when the displacement from equilibrium is a maximum, the velocity is zero When x = +A and x = -A, object is momentarily at rest The speed is a maximum when the displacement from equilibrium is zero, i.e. vmax = Aω (since largest value of sinθ =1) 5. Oscillations About Equilibrium 7 ACCELERATION IN SIMPLE HARMONIC MOTION • • • • • • • • • The acceleration of an object in uniform circular motion is given by acp = rω2 Direction of acceleration: toward centre of circular path When the angular position of the peg is θ, the acceleration vector is at an angle θ below the x-axis, and its x component is –acpcosθ Setting r = A and θ = ωt the acceleration in simple harmonic motion is a = -Aω2cos(ωt) m/s2 Position and acceleration for SHM are plotted below Acceleration and position vary with time in the same way, but with opposite signs When the position has a maximum positive value, the acceleration has a maximum negative value After all, the restoring force is opposite to the position, hence the acceleration a = F/m must also be opposite to the position, thus a = -ω2x Maximum acceleration occurs when –cosθ = 1, thus the maximum acceleration amax = Aω2 5. Oscillations About Equilibrium 8 SIMPLE HARMONIC MOTION: EXAMPLE • A 747 is hit by turbulence. Upon landing, data from the airliner’s black box indicates that the plane moved up and down with an amplitude of 30.0m, and a maximum acceleration of 1.8g. Treating the up and down motion of the plane as simple harmonic, find the time required for one complete oscillation. Also find the plane’s maximum vertical speed. 5. Oscillations About Equilibrium 9 THE PERIOD OF A MASS ON A SPRING • • • • • • • • • Net force exerted by a spring with constant k, acting on a mass m at position x is F = -kx Since F = ma, it follows that ma = -kx Substituting: m[-Aω2cos(ωt)] = -k[Acos(ωt)] Which yields ω2 = k/m or ω = √(k/m) Noting that ω = 2π/T, it follows that the period of a mass on a spring is T = 2π√(m/k) seconds The period increases as the mass increases, and decreases with the spring’s force constant A larger mass has greater inertia, and thus takes longer for the mass to move back and forth through an oscillation A larger value of the force constant k indicates a stiffer spring, which means a mass would complete an oscillation is less time that it would on a softer spring Example: A 0.120kg mass attached to a spring oscillates with an amplitude of 0.075m and a maximum speed of 0.524 m/s. Find the force constant and the period of motion. 5. Oscillations About Equilibrium 10 A VERTICAL SPRING • • • • • • • Up to now, we have considered springs that are horizontal When a mass m is attached to a vertical spring, it causes the spring to stretch The vertical spring is in equilibrium when it exerts an upward force equal to the weight of the mass, i.e the spring stretches by an amount y0 Thus, for vertical springs: ky0 = mg Or y0 = mg/k A mass on a spring oscillates around the equilibrium point y = y0, and also exhibits simple harmonic motion Example: A 0.26kg mass is attached to a vertical spring. When the mass is put into motion, its period is 1.12s. How much does the mass stretch the spring when it is at rest in its equilibrium position? 5. Oscillations About Equilibrium 11 THE SIMPLE PENDULUM • • • • • A simple pendulum consists of a mass m suspended by a light string or rod of length L The pendulum has a stable equilibrium when the mass is directly below the suspension point, and oscillates about this position if displaced from it Consider the potential energy of the system When the pendulum is at an angle θ with respect to the vertical, the mass m is above its lowest point by a vertical height L(1 – cosθ) If the potential energy is 0 at θ = 0, then the potential energy for general θ is: U = mgL(1 – cosθ) 5. Oscillations About Equilibrium 12 FORCES ACTING ON A PENDULUM • • • • • • • • • • • The forces acting on the mass are due to gravity and the tension in the string (both are vectors) The tension acts in the radial direction, and keeps the mass moving along its circular path The net tangential force acting on m is just the tangential component of its weight: F = mgsinθ The direction of the net tangential force is always toward the equilibrium point, thus F is a restoring force For small angles sinθ ≈ θ, so the arc length s = Lθ Equivalently θ = s/L If the mass is displaced from equilibrium by a small arc length s, the force is F = mgsinθ ≈ mgθ = (mg/L)s Comparing to a mass on a spring, F = kx Letting x = s and k = mg/L Period of a pendulum is the same as that of a mass on a spring T = 2π√(m/k) = 2π√(m/(mg/L)) = 2π√(L/g) seconds 5. Oscillations About Equilibrium 13 PENDULUM EXAMPLE • A pendulum is constructed from a string 0.627m long attached to a mass of 0.25kg. When set in motion, the pendulum completes one oscillation every 1.59s. If the pendulum is held at rest and the string is cut, how long will it take for the mass to fall through a distance of 1.0m? 5. Oscillations About Equilibrium 14 DAMPED OSCILLATIONS (1) • • • • • • • • • • Up to this point, oscillating systems in which no mechanical energy is gained or lost (due to friction, air resistance) have been considered In most physical systems, there are energy losses As the mechanical energy of a system decreases, its amplitude of oscillations decrease as well This type of motion is called a damped oscillation An oscillating mass may lose its mechanical energy to a force such as air resistance that is proportional to the speed of the mass r and r opposite in direction, which can be defined as F = −bv The damping constant is b and is a measure of the strength of the damping force (kg/s) If the damping constant is small, the system will continue to oscillate, but with a continuously decreasing amplitude – called underdamped motion In underdamped motion, the amplitude decreases exponentially with time: A = A0e-bt/2m where A0 is the initial amplitude, m is the mass, time t As the damping is increased, a point is reached where the system no longer oscillates, but simply relaxes back to the equilibrium position – critically damped The system is overdamped if the damping is increased beyond this point, and the system still returns to equilibrium without oscillating, but takes longer 5. Oscillations About Equilibrium 15 DAMPED OSCILLATIONS (2) Shock absorbers are designed to be critically damped, so that when a car hits a bump, it returns to equilibrium quickly without oscillations 5. Oscillations About Equilibrium 16 DRIVEN OSCILLATIONS • • • • • • • • • • Applying an external force to a system furnishes it with energy – positive work is done If you hold the end of a string from which a small weight is suspended, and the weight is set in motion, holding your hand still will cause it to stop oscillating If you move your hand back and forth, the weight will oscillate indefinitely This motion is called a driven oscillation Moving your hand at the wrong frequency (too high or too small) will cause the weight to oscillate erratically Oscillating your hand at an intermediate frequency will result in large amplitude oscillations for the weight This frequency is called the natural frequency, f0, of the system Natural frequency of a pendulum: f0 = 1/T = (1/2)√(g/L) For a mass on a spring: f0 = 1/T = (1/2)√(k/m) Generally driving any system near its natural frequency results in large oscillations 5. Oscillations About Equilibrium 17 RESONANCE • • • • • • • • Consider a plot of amplitude, A, versus driving frequency f for a mass on a spring There are large amplitudes for frequencies near f0 This type of large response is known as resonance The curves are referred to as resonance curves Systems with small damping have a high narrow peak on their resonance curve Tuning a radio changes the resonance frequency of an electrical circuit (made up of capacitors, inductors and resistors) in the tuner When its resonance frequency matches the frequency being broadcast by a station, that station is picked up Bridges affected by high winds may swing with large amplitudes, especially if the frequency of the wind matches the natural frequency of the bridge as a system (recall the Millennium Bridge fiasco) 5. Oscillations About Equilibrium 18