MECHANICS IV  SIMPLE HARMONIC MOTION


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1 MIVp.1 A. OSCILLATIONS B. SIMPLE PENDULUM C. KINEMATICS OF SIMPLE HARMONIC MOTION D. SPRINGANDMASS SYSTEM E. ENERGY OF SHM F. DAMPED HARMONIC MOTION G. FORCED VIBRATION A. OSCILLATIONS A toandfro repeating motion is called an oscillation or vibration. The following are examples of some mechanical oscillations. [ Advanced Physics Duncan 3 rd Ed John Murray 1987 p.186] Every oscillation is specified by its: period (T) amplitude (A) frequency (f) or angular frequency (ω = 2πf). Oscillation is driven by restoring force. Restoring force always point at a central point. The central point of oscillation is called the equilibrium position. Q In the above examples of oscillation, point out the origin of restoring force and find where the equilibrium position is. Chapter 8  p.1
2 MIVp.2 B. SIMPLE PENDULUM Some kinds of oscillation obey the force law F x. They are called simple harmonic motion. θ x l T mg A simple pendulum is a typical example of SHM. This simplest form is just a metal bob attached to the end of a string. On swinging, the bob can only move along the tangential direction of the string. For small angles of oscillation (θ < 10 ), we can assume the bob moves along the horizontal direction (xdirection). 1. Analysis of motion x A x = max, v = 0, a = max x = 0, v = max, a = 0 T 2T t A x = max, v = 0, a = max CW Draw the corresponding vt and at graphs for the above SHM. Chapter 8  p.2
3 MIVp.3 2. Dynamic treatment Resolving force components along (x / tangential) and perpendicular (y / radial) to direction of motion: ydirection: no motion T = mg cosθ T xdirection: F = ma mg sinθ = ma x Note that the net force is always in the opposite direction to the displacement. This force tries to keep the bob back to the equilibrium position and is called a restoring force. For small angle θ, we use the approximation x l θ l sinθ sinθ x/l Thus, mgx/l = ma a = (g/l) x This is a differential equation in the form 2 d x = ω 2 x : 2 dt General solution: x = A cos(ωt + φ) where A and φ are constants If the bob is released at t = 0, matching the initial conditions x = A, solution x = A cosωt angular frequency ω = (g/l) period of oscillation T = 2π/ω = 2π (l/g) Note that the period of oscillation does not depend on the amplitude of oscillation. This is called isochronous. The point x = 0 is called equilibrium position. At this point, there is no net force on the bob along the direction of motion. mg Chapter 8  p.3
4 MIVp.4 Q A pendulum is designed to measure time. The simplest way is to make it oscillate at 1 s per cycle. What is the length of this pendulum? Find the value of ω and thus write down the xt relationship. Complete the table for this pendulum. Suppose A = 10 cm, and the pendulum is released at t = 0. t /s ωt / rad ωt / x / cm π Q Why isochronous oscillation so important in our daily life? Give examples. Q from x = A (one end) to: shortest time needed t A (the other end) ½ T 0 (equilibrium position) ¼ T ½ A (halfway to equilibrium position)? Chapter 8  p.4
5 MIVp.5 C. KINEMATICS OF SIMPLE HARMONIC MOTION 1. Definition Simple harmonic motion is oscillation in which acceleration is directly proportional to displacement and in opposite direction to the displacement. Written in equation, a x 2. Equations of motion Q x = v = dx/dt = a = dv/dt = Eliminating t, a = ω 2 x v = ±ω (A 2 x 2 ) A cosωt ωasinωt ω 2 Acosωt definition of SHM Is x = A sin ωt represents a SHM? What is the difference compared with x = A cos ωt? CW [Ch.8 Q.4] A pendulum of length 2.0 m makes small angle oscillations with an amplitude of 15. (a) Find the time required for the bob to oscillate from 5 to 10 towards the right. (b) Calculate the velocity and acceleration at these two positions. Chapter 8  p.5
6 MIVp.6 3. Rotating vector model Simple harmonic motion can be regarded as the projection of uniform circular motion. circular motion θ = ω t v = ωa a = ω 2 A f = ω / 2π projection on xaxis x = A cosωt v x = ωa sinωt a x = ω 2 A cosωt f = ω / 2π y A θ x a v x r = xiˆ + yj ˆ = Acosθ iˆ + Asinθˆj = Acosωtiˆ + Asinωtj ˆ v = r = a = v = d dt d dt d ( xiˆ + yj ˆ) = ( Acosω tiˆ + Asinωtj ˆ) = ωasinωtiˆ + ωacosωtj ˆ dt A tiˆ 2 A tj A tiˆ 2 ( ω sinω + ω cosω ˆ) = ω cosω ω Asinωtj ˆ (xcomponent of circular motion is SHM) EXPT Visualising the rotating vector model [ Advanced Physics Duncan 3 rd Ed John Murray 1987 p.188] 4. Phase relationship From the xt, vt and at graph of SHM, we see that a attains its maximum first, followed by v after ¼ T, and x another ¼ T later. We say that acceleration leads velocity by 90 (½ π) and velocity leads displacement by 90 (½ π). The opposite word for lead is lag. Q Acceleration leads displacement by. Displacement velocity by. Chapter 8  p.6
7 MIVp.7 D. SPRINGANDMASS SYSTEM Hooke s law describes the restoring force supplied by a spring: F = k x k is the spring constant in N m 1 1. Mass hung vertically on a spring l kl x k(l+x) an unloaded spring mg A mass is hung on the end of the spring and the spring is stretched to a length of l longer. kl = mg mg F The mass is pulled further downwards by a displacement x. F + mg = k (l + x) F + mg = k l + kx kl = mg applied force: F = k x When the mass is released, it oscillates up and down under the restoring force. restoring force (net force) F = k x F = ma kx = ma a = (k/m)x Solution: x = angular frequency: period of oscillation: ω = T = Chapter 8  p.7
8 MIVp.8 CW A spring of spring constant 10 Nm 1 is loaded with a 100 g mass. It is pulled down by 3 cm and released. Find the period, amplitude, maximum speed and maximum acceleration of the oscillation. CW [Ch.8 Q.9] (a) A block of mass 0.2 kg is hung from a spring of force constant 10 N m 1. What is the period of the SHM? (b) If the amplitude of vibration is A, find the maximum velocity and maximum acceleration in terms of A. (c) An ant sits on top of the block. What must be the amplitude of vibration if the ant is to be shaken from the block at least momentarily? At which point of the cycle does this occur? Chapter 8  p.8
9 MIVp.9 2. Springs in series and parallel [ Advanced Level Physics Nelkon & Parker 6 th Ed Heinemann 1987 p.8486] Chapter 8  p.9
10 MIVp.10 Chapter 8  p.10
11 MIVp.11 E. ENERGY OF SHM An oscillation system consists of two elements: elasticity to stored potential energy and mass to have kinetic energy. From the view of energy, oscillation is actually the continuous exchange between potential energy and kinetic energy. For example, a simple pendulum leaving its highest point have all energy stored as potential. As it falls, its kinetic energy increases. Kinetic energy attains a maximum as it passes through the equilibrium (lowest) position and then changing to potential energy after passing that point, until all kinetic energy has transformed to potential energy at the other end. This process repeats and repeats. U energy E=K+U K position Consider a springandmass system E k = ½ mv 2 = ½ mω 2 (A 2 x 2 ) E p = ½ F x = ½ kx 2 = ½ mω 2 x 2 Total energy E = E k + E p E = ½ mω 2 A 2 2 = ½ mv max or E = ½ ka 2 2 = ½ kx max At x = A, v = 0, E k = 0, E p = E At x = 0, v = max, E k = E, E p = 0 Energy as function of time: E k = ½ mv 2 E p = ½ mω 2 x 2 = ½ mω 2 A 2 sin 2 ωt = ½ mω 2 A 2 cos 2 ωt CW [Ch.8 Q.10] An object of mass 0.1 kg oscillates in SHM of amplitude 0.05 m. If the total mechanical energy of the object is J, find the period of its motion. Chapter 8  p.11
12 MIVp.12 F. DAMPED HARMONIC MOTION 1. No Damping The oscillation will go on indefinitely if there is no energy loss. 2. Underdamped Oscillation When a pendulum is allowed to swing in air, its amplitude dies away slowly and finally comes to rest. Energy of the pendulum is lost slowly due to air resistance. The equation of motion is now x = A e γt/2 cosω 1 t Comparing with x = A cosωt, The new amplitude A e γt/2 shows the amplitude is decreasing exponentially, and the vibration becomes slower due to the damping force, the angular frequency is replaced by a smaller ω Overdamped Oscillation In this case, the damping force is so large that there is no oscillation at all. For example, a mass hung on a spring is put into oil. The only motion is the mass getting back to its equilibrium position slowly. 4. Critically Damped Oscillation This is between the cases of slightly damped and overdamped. The oscillator goes back to equilibrium position quickly without any oscillation. Chapter 8  p.12
13 MIVp.13 G. FORCED VIBRATION A forced vibration is an oscillation under an external vibrating force. Although the driving frequency ω may or may not match the natural frequency ω 0 of the oscillator, the oscillator is forced to vibrate at the same frequency as the driver. DEMO Barton s pendulum [ Advanced Physics Duncan 3 rd Ed John Murray 1987 p.194] 1. Resonance ω = ω 0 The pendulum with length the same as the driver vibrates with the greatest amplitude. Its natural frequency is equal to the driving frequency of the oscillator. This shows that energy transfer is most efficient at resonance. 2. Low driving frequency ω << ω 0 The short pendulums (of fast natural frequency) vibrate in phase with the driver. Their amplitudes are obviously smaller. 3. High driving frequency ω >> ω 0 The long pendulums vibrate exactly (180 ) out of phase as the driver. Again, their amplitudes are small. T W CHAN Oct 2001 Chapter 8  p.13
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