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Transcript
1 Handout 5: Oscillatory motion Simple harmonic motion Simple harmonic motion (SHM) is periodic motion with constant period or frequency. It is the motion where acceleration is proportional to displacement in the opposite direction. The acceleration π and displacement π₯ are related by π = βπ2 π₯, Figure 1: Mass-spring system where π is the angular frequency of the motion. The period and frequency can be found from π. In Figure 1, a mass π on a smooth floor is attached to one end of a spring whose other end is fixed. The spring has spring constant π. The mass is displaced to the right and let go; it performs an oscillatory motion along x-axis. When the displacement is π₯, there is a force β ππ₯ acting on the mass according to Hookeβs law. This force in the spring is a restoring force. The equation of motion is ππ₯ = βππ₯ β π₯ = β It can be seen that π2 = π/π or π = the period of oscillations π = π π₯. π π/π. Therefore, 2π π = 2π . π π Example A mass π is suspended from a vertical spring whose other end is fixed. The spring constant is π. Show that the period of motion is π = 2π π . π Example A mass π is attached by two springs with different spring constants π1 and π2 . Show that the mass undergoes SHM and find the formula for the period of the motion. 2 Example A wooden block of mass π and cross-section area π΄ is partially immersed in a liquid with density π. The block is displaced downwards by a distance and released. Show that the block performs SHM and determine the angular frequency of the motion. Solution to SHM equation Letβs go back to the mass-spring system as shown in Figure 1. The mass is displaced by distance π΄ and let go. Therefore, π΄ is the maximum displacement which is known as the amplitude of the oscillations. The equation of motion is π2 π₯ = βπ2 π₯, ππ‘ 2 π = π . π We wish to find displacement π₯(π‘) and velocity π£(π‘) as functions of time π‘. The general solution is π₯ π‘ = πΆ1 sin ππ‘ + πΆ2 cos ππ‘, where πΆ1 and πΆ2 are constants. The above equation can be expressed in the form π₯ π‘ = π΄ sin(ππ‘ + πΏ), where π΄ is the amplitude and πΏ is a phase angle. The shape function π₯ π‘ is illustrated in Figure 2. There are two cases to consider ο· At π‘ = 0, π₯ = 0, we find that πΏ = 0: π₯ π‘ = π΄ sin ππ‘. ο· At π‘ = 0, π₯ = π΄, we find that πΏ = π/2: π₯ π‘ = π΄ cos ππ‘. From displacement π₯ π‘ = π΄ sin(ππ‘ + πΏ), one can differentiate to obtain velocity π£(π‘) and acceleration π π‘ : Figure 2: Displacement against time 3 π£ π‘ = ππ΄ cos(ππ‘ + πΏ) , π π‘ = βπ2 π΄ sin(ππ‘ + πΏ) The graphs of π£(π‘) and π π‘ are shown in Figure 3. It should be noted that the maximum velocity π£πππ₯ and maximum acceleration ππππ₯ are given by π£πππ₯ = ππ΄, ππππ₯ = π2 π΄. Example A particle performs simple harmonic motion where acceleration π in ms-2 is a function of displacement π₯ in m according to π = β2.65π₯. The amplitude of the motion is 0.35 m and the particle is at π₯ = 0 when time π‘ = 0 s. Determine a) Angular frequency b) Maximum speed c) Displacement when π‘ = 2.5 s Example For a SHM motion with solution π₯ π‘ = π΄ sin(ππ‘ + πΏ), show that velocity π£(π₯) as a function of π₯ is given by π£ = π π΄2 β π₯ 2 . Example A particle π moves along the π₯-axis. At time π‘ seconds, its displacement, π₯ meters, from the origin π is given by π₯ = 5 sin π 3 π‘ . Points π΄ and π΅ are on π₯-axis such that ππ΄ = 2 m and ππ΅ = 3 m. Find the time taken by π to travel directly from π΄ to π΅. Figure 3: Displacement, velocity and acceleration against time 4 Example A particle P moves on the x-axis with simple harmonic motion about the origin O as centre. When P is a distance 0.04 m from O, its speed is 0.2 msβ1 and the magnitude of its acceleration is 1 msβ2. The amplitude of the motion is A meters. a) Find the period of oscillations. b) Find the amplitude A. c) Find the total time, within one complete oscillation, that distance OP is greater than A/2. Damped harmonic motion For simple harmonic motion, the total energy of the system is constant; the amplitude of the oscillations is also constant. If damping force is taken into account, the energy is lost and therefore the amplitude decreases with time. Such motion is called damped harmonic motion. The damped harmonic motion can be modeled as shown in Figure 4. Apart from the usual SHM setup which consists of mass and spring, the mass is connected to a light vane immersed in a liquid. As the mass moves at speed π₯, there is a drag force β ππ₯ acting on the vane. The constant π is related to the viscosity of the liquid. The equation of motion for the mass becomes ππ₯ = βππ₯ β ππ₯. Figure 4: Damped harmonic motion setup For b not too large, the solution to the above equation is given by π₯ π‘ = π΄0 π β(π 2π )π‘ sin ππ‘ + πΏ , π = π π2 β , π 4π2 where π΄0 is a constant. The value of b has to be small enough to keep π real. The term in front of sine function represents the amplitude A of the motion which is decreasing with time (Figure 5): π΄ = π΄0 π β(π 2π )π‘ . Hence, the constant π΄0 is the initial amplitude. This type of damping is called underdamping. If the value of π is Figure 5: Decay in amplitude of underdamped motion. 5 too large, the damping may become critical damping or overdamping. The comparison is illustrated in Figure 6. For critically damped motion, the value of b is high, there is no oscillation and π₯ reaches zero in short time. For overdamped motion, the value of b is even higher; there is also no oscillation and π₯ reaches zero in a long time. Example A particle of mass π = 0.1 kg is attached to a spring with constant π = 250 Nm-1 and undergoes a damped harmonic motion with drag force with coefficient π = 0.1 kg s-1. a) Evaluate the period of the oscillations. b) Find time at which the amplitude is halved. c) By the time in b), how many cycles have passed? Figure 6: Types of damped harmonic motion Example For a damped harmonic motion of an object with drag force π = βππ£ where π is a constant and π£ is the speed of the object, show that the rate of change in energy of the system is equal to βππ£ 2 . Forced oscillations and resonance Mass-spring system has a natural frequency π0 = 1 π , 2π π (assume no damping). That means whenever the mass oscillates, it will do so with the natural frequency. In Figure 7, the mass is forced to oscillate by an oscillatory driving force by hand on the top end of the spring. The driving force has a driving frequency π. If the driving frequency is equal to the natural frequency, the result is Figure 7: Driven oscillations. The mass is forced to oscillate. 6 that the mass will oscillate with large amplitude. This phenomenon is known as resonance. At resonance, the amplitude of the oscillation is large. The size of the amplitude at resonance depends on the degree of damping as shown in Figure 8. It is important to look at the phase difference πΏ between driving force (hand) and the oscillating mass. ο· When π βͺ π0 , we find πΏ = 0 ο hand and mass are in phase. ο· When π = π0 , we find πΏ = π/2 ο hand and mass out of phase by π/2. ο· When π β« π0 , we find πΏ = π ο hand and mass out of phase by π. Resonance occurs in other systems too. A system has its own natural frequencies. When the frequency of the driving force matches one of the natural frequencies, resonance occurs and it can be destructive. Examples are ο· Glass-breaking sound The frequency of sound matches the natural frequency of the glass, the result is strong vibrations of the glass until it breaks. ο· Loud speakers Boom or buzz is heard when a musical note happens to coincide with the resonant frequency of the speaker cone. ο· Tuned radio A tuned circuit in radio receiver responds strongly to a particular frequency. ο· Airplaneβs wings If frequency of vibrations of the engines of particular airplane is right to resonate with the natural frequencies of the wings. Large oscillations build up and the wings may fall off. ο· Tacoma Narrows Bridge (Figure 9) The wind across the bridge was turbulent and vortices were formed with regular frequency. This frequency matched the natural frequency of the bridge. The result was violent vibrations of the bridge which collapsed eventually. Figure 8: Graphs of amplitude against driving frequency. Figure 9: Tacoma Narrows Bridge