(c) R Boasman 2006 Physics 3.4 Demonstrate understanding of mechanical systems Credits: 6 This achievement standard involves knowledge and understanding.

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(c) R Boasman 2006 Physics 3.4 Demonstrate understanding of mechanical systems Credits: 6 This achievement standard involves knowledge and understanding of phenomena, concepts, principles and relationships related to translational, circular and rotational motion, and simple harmonic motion, and the use of appropriate methods to solve related problems.

(c) R Boasman 2006 Achievement Criteria AchievementMeritExcellence Identify or describe aspects of phenomena, concepts or principles. Give descriptions or explanations in terms of phenomena, concepts, principles and/or relationships. Give concise explanations, that show clear understanding, in terms of phenomena, concepts, principles and/or relationships. Solve straightforward problems where the concept will be obvious and the solution involves a single process. Solve problems where the relevent concept is not immediately obvious and may involve rearranging a formula. Solve complex where two different concepts are needed and the solution involves more than one process.

(c) R Boasman 2006 Translational Motion Centre of mass (1 and 2 dimensions); Conservation of momentum and impulse (2 dimensions only).

(c) R Boasman 2006 Circular and Rotational Motion Velocity and acceleration of, and resultant force on, objects moving in a circle under the influence of 2 or more forces, eg banked corners, vertical circles; Newton’s Law of gravitation, satellite motion. Rotational motion with constant angular speed and with constant angular acceleration; Torque; Rotational inertia; Angular momentum; Rotational kinetic energy; Conservation of angular momentum; Conservation of energy.

(c) R Boasman 2006 Simple Harmonic Motion (SHM) Displacement; velocity; acceleration; time and frequency of a particle undergoing SHM; Forced SHM; Resonance; The reference circle; Phasors; Conservation of energy.

(c) R Boasman 2006 Relationships E P = ½ky 2

(c) R Boasman 2006 This is motion in continually repeating cycles. Two examples are Circular motion and Oscillatory motion. Circular motionOscillatory motion r = radius (m), v = linear velocity (ms -1 ), ω = angular velocity/frequency (rad s -1 ) T = Time period for one cycle (s), f = frequency (Hz), Periodic Motion v= 2 πr T r ω Equilibrium position θ v T = 1 f T = 2π ω ω = 2πf r

(c) R Boasman 2006 Simple Harmonic Motion SHM is a periodic motion where the oscillations are repeated again and again. For example: a Pendulum, a Mass on a spring, a piston in a car engine, a guitar string, an electron in a wire carrying alternating current, a boat bobbing on the water, the changing tides.

(c) R Boasman 2006 An object exhibits SHM if … “The force F on the object acts towards the equilibrium position and is proportional to the distance, y from the point” - F+ F + y- y + y- y + F - F Equilibrium position The relationship is linear and as Y = MX + C so: F = -ky by Newton’s 2 Law F = ma hence: a = -ky m

(c) R Boasman 2006 Linking Circular motion and SHM θ ω Particle radius, A. y = 0 y max = +A Displacement y max = -A y = A sinθ Velocity v max = v lin So v max = Aω v = 0 ms -1 v = Aω cosθ θ v lin Acceleration a = 0 ms -2 a max = - Aω 2 a max = + Aω 2 a = - Aω 2 sinθ + - A θ Circular motion SHM “shadow”

(c) R Boasman 2006 Displacement θ ω Equilibrium position y = 0 y = A sin ωt time, t + - A Displacement t = θ ω

(c) R Boasman 2006 Velocity θ ω Equilibrium position time, t + - v = A ω cos ωt v Velocity t = θ ω A

(c) R Boasman 2006 Acceleration θ ω Equilibrium position a = 0 a = - A ω 2 sin ωt time, t + - Acceleration t = θ ω

(c) R Boasman 2006 Phasor Diagrams All three phasors can be drawn on a single diagram. Phasors rotate anticlockwise with angular velocity, ω. Acceleration phasor leads the velocity phasor by π/2 (90˚) and the displacement phasor by π (180˚). Velocity phasor leads the displacement phasor by π/2 (90˚). Displacement phasor Acceleration phasor Velocity phasor A Aω Aω 2 ω

(c) R Boasman 2006 Phasor equations For an oscillating body starting from the equilibrium position, y = 0, and moving in a positive direction at time t = 0 s, the following equations apply. y = A sin ωt v = Aω cos ωt a = - Aω 2 sin ωt = - ω 2 y For an oscillating body starting at maximum amplitude, y = A (i.e. one end of its motion) and moving in a positive direction at time t = 0 s, the following equations apply. y = A cos ωt v = - Aω sin ωt a = - Aω 2 cos ωt

(c) R Boasman 2006 Time Period of oscillatory motion As a = - ω 2 Asin ωt and y = Asin ωt So a = - ω 2 y For a particle traveling in a circle T = 2π ω hence;T = 2π But a = - ky Combining gives us: m T = 2π

(c) R Boasman 2006 Simple pendulum The SHM is due to the force F , where: F  = F g sinθ As F = ma soma = mg sinθ, hence a = g sinθ But for very small angles sin θ = y/l, hence a = gy towards the equilibrium position. l a = - gyin the direction that y increases. l This is of the from a = -ky/m therefore SHM. As T = 2π/ωT = 2π√l/g Equilibrium position θ FgFg FF F║F║ r T y l

(c) R Boasman 2006 Conservation of energy. Consider a mass sitting on a frictionless surface and attached to a spring. If the spring is stretched and then released the mass will begin to oscillate in SHM. During SHM the total energy of the system remains the same in agreement with the principle of conservation of energy. However the form the energy takes is continually changing from potential to kinetic and back again. When the mass is in any other position it has both potential and kinetic energy. The kinetic energy at any point is given by the formula; E K = ½ mv 2.

(c) R Boasman 2006 Potential Energy At some point from the equilibrium position with displacement, y, the potential energy is equal to the work done, W, in moving the mass from the equilibrium position to the other position. Work done = Force × distance moved in the direction of the force However the force is not constant and increases as the distance from the equilibrium position increases. y F

(c) R Boasman 2006 Potential Energy As y increases so does the force. For a spring F = ky (As this equation refers to a force increasing in the same direction as the displacement it is not negative) Hence to find the work done, the area under a graph of force against distance must be found. Work done = Area under graph Work done = ½ ky 2 As Work done = Potential energy gained Hence E P = ½ ky 2 distance y Force ky

(c) R Boasman 2006 Total energy The total energy of the oscillating object will be the sum of the energies at any point. E total = E P + E K HenceE total = ½ ky 2 + ½ mv 2 At maximum amplitude, y = A and v = 0, hence E K = 0 J, so E total = ½ kA 2 And since total energy is conserved, the oscillating object will always have this amount of energy. ½ kA 2 = ½ ky 2 + ½ mv 2 ½ kA 2 Energy time0 Energy T/4T/23T/4T EPEP EKEK EPEP EKEK 0 - AA y

(c) R Boasman 2006 Damping An object exhibiting SHM will experience frictional forces acting on it. With friction acting on the body energy is slowly lost from the system and the resultant motion is no longer true SHM. With time the amplitude of each oscillation decreases. Damping is often unhelpful and would cause pendulum clocks to stop working due to the energy lost from the pendulum with each swing. In car suspension systems, damping is used to eliminate unwanted oscillations after a car goes over a bump in the road. Without the damping the springs would cause the car to keep bouncing forever. y t y t Light damping Heavy damping

(c) R Boasman 2006 Critical damping When damping exceeds a certain critical value the d-t graph no longer has a sinusoidal shape bounded by an exponential decay envelope. Instead the graph itself has an exponential shape and no oscillation occurs. y t Very heavy damping

(c) R Boasman 2006 Forced SHM All bodies have a natural frequency of oscillation (f 0 ) that occurs when the body is displaced from its equilibrium position and then released. If a body is subjected to a periodic external force (driving force), then it will undergo forced SHM. The frequency of the forced SHM will equal the frequency of the external force (driving frequency, f D ).

Tacoma Narrows Disaster (c) R Boasman 2006

Resonance The amplitude of the forced SHM is small unless the driving frequency is close to the natural frequency. When f D ≈ f 0 the amplitude of the forced SHM becomes significant. The frequency at which the forced SHM has the maximum amplitude is known as the resonant frequency (f R ). With no damping the resonant frequency is the same as the natural frequency. As damping increases the resonant frequency decreases. Driving frequency 0 Amplitude of forced SHM oscillations Light damping Heavy damping fOfO fOfO

(c) R Boasman 2006 Consequences of SHM Soldiers need to break step when crossing a bridge. Opera singers can shatter glasses by forcing them to vibrate at their natural frequency. A diver on a spring board builds up the amplitude of oscillation of the board by ‘bouncing’ on it at its natural frequency. Loose parts in cars will rattle at certain speeds. A column of air can be made to resonate to a particular note. Electrical resonance is used to tune radios. Resonant vibrations of quartz crystals is used to control clocks and watches.

(c) R Boasman 2006 Driving pendulum Barton’s pendulum A heavy driving pendulum is displaced so that it oscillates at its natural frequency (into and out of the page). It’s vibrations are passed through the connecting string to the other pendulums and they are forced to oscillate. All the pendulums oscillate with the same frequency as the driving pendulum. The pendulum with the same length as the driving pendulum has the same natural frequency so oscillates with the largest amplitude (resonance), and is ¼ of a period behind the driving pendulum. The shorter pendulums are in phase with the driving pendulum. The longer pendulum is ½ a period behind.

(c) R Boasman 2006 Forced SHM and resonance. When the driving frequency is low, the forced SHM is in phase with the driving force. When the driving frequency is much higher than the natural frequency, the forced SHM is out of phase with the driving force. π2π2 π Driver frequency f0f0 Forced oscillation is T/2 out of phase. Forced oscillation is T/4 out of phase. Forced oscillation is in phase. Phase difference