# Lecture12 - Oscillatory Mo3on Periodic and simple...

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Unformatted text preview: Oscillatory Mo3on Periodic and simple harmonic mo3on Oscilla'on or periodic mo'on: mo3on that repeats itself, e.g. a mo3on of a swing, pendulum of a grand father clock, vibra3on of a guitar string. Mo3on repeats because of some restoring force ac3ng on the body in mo3on. Chapter 15 Simple harmonic mo'on is a special kind of mo3on when the restoring force is propor3onal to the posi3on of the object rela3ve to some equilibrium posi3on. Spring ­mass system: a model for simple harmonic mo3on Simple harmonic mo3on = restoring force on the object is propor3onal to the posi3on of the rela3ve to some equilibrium posi3on. A good example of such a force: the spring force (Hook’s law) F =  ­ kx d2x Using Newton’s second law: m 2 = − kx dt d2x k ⇒ 2 = − x = −ω 2 x dt m a = −ω 2 x where ω= k m Thus a body in simple harmonic mo3on has an accelera3on propor3onal to the posi3on and oppositely directed to the to the displacement from equilibrium. Par3cle in a simple harmonic mo3on Equa3on of mo3on: d2x = −ω 2 x dt 2 ω= k m Solu3on to this diﬀeren3al equa3on: x (t ) = A cos(ω t + ϕ ).....(1) ω = angular frequency A = amplitude φ = phase constant (ini3al phase angle) ωt + φ = phase of the mo3on Equa3on (1) shows that the par3cle in simple harmonic mo3on is a periodic mo3on. If T is the 'me period, the 3me interval in which the par3cle goes through one full cycle of mo3on, then 2π [ω (t + T ) + ϕ ] − (ω t + ϕ ) = 2π ⇒ ω T = 2π ⇒ ω = T Frequency of oscilla3on is the number of full cycles completed in per unit 3me (second): 1 ω f= ⇒f= → ω = 2π f T 2π SI unit of frequency is hertz (Hz). 1 Hz = 1 cycle per second. Par3cle in a simple harmonic mo3on (contd…) From the expression of posi3on of a body in simple harmonic mo3on x (t ) = A cos(ω t + ϕ ), we can derive the expressions for velocity and accelera3on of the par3cle: v(t ) = dx ⇒ v(t ) = −ω A sin(ω t + ϕ ) dt dv ⇒ a(t ) = −ω 2 A cos(ω t + ϕ ) dt Also, vmax = ω A a(t ) = amax = ω 2 A Par3cle in a simple harmonic mo3on (contd…) a v F,a v a v Example A block with mass m = 0.68 kg is fastened to a spring with k=65 N/m. The block is pulled a distance x = 11 cm from its equilibrium posiIon at x = 0 on a fricIonless surface and released from rest. (a) What are the angular frequency, frequency, and the period of the resulIng moIon ? (b) What are the amplitude, maximum speed and maximum acceleraIon ? (c) What is the displacement funcIon x(t) for the moIon ? (a) (b) Amplitude = maximum displacement. Since the block starts from rest it can only return to a maximum deﬂec3on of 11 cm, therefore the amplitude A = 11 cm. Maximum speed: vm = ω A = (9.78 rad/s)(0.11m ) = 1.1m/s am = ω 2 A = (9.78 rad/s)2 (0.11m ) = 11m/s 2 Maximum acceleraIon: (c) Displacement func3on: x (t ) = A cos (ω t + ϕ ) At 3me t = 0, x = A, therefore ⇒ ϕ = 0 ⇒ x(t ) = A cos (ω t ) = 0.11 cos(9.8t ) What is the value of ϕ ? x(0 ) = A cos ϕ = A ⇒ cos ϕ = 1 where displacement in meters, 3me in seconds (SI units) Energy and simple harmonic mo3on Poten3al energy at any point in 3me: U (t ) = 1212 kx = kA cos 2 (ω t + ϕ ) 2 2 Kine3c energy at any point in 3me (use k = mω2 and v(t) = -ωA sin(ωt+ϕ)): 12122 K (t ) = mv = kA sin (ω t + ϕ ) 2 2 Total energy: E = K +U = 12 kA [cos 2 (ω t + ϕ ) + sin 2 (ω t + ϕ )] 2 1 E = U + K = kA2 2 Example At Ime t = 0 a spring pendulum with k = 12 N/m and m = 0.5 kg is moving with velocity v = - 0.7 m/s at a displacement of x = 8 cm. What is the pendulums total energy and amplitude? We know: x(0 ) = A cos ϕ = 0.08 m v(0 ) = −ω A sin ϕ = −0.7 m/s Need to ﬁnd ϕ: Amplitude: Energy: ⇒A= x (0 ) 0.08 m = = 0.164 m cos ϕ cos 60.75° 121 kA = (12 N/m )(0.164 m )2 = 0.161J 2 2 E = K +U = Pendulums The spring pendulum Restoring force: Then: T= 2π ⇒ T = 2π ω m k The torsion pendulum As the wire twists, there is a restoring torque: Instead of mass we have to use the rota3onal iner3a. We ﬁnd: The physical (real) pendulum The restoring torque is: where h = distance from pivot to CG Newton’s second law: Solve for α: (Compare to Then: and: for a spring pendulum.) The simple pendulum Special case of physical pendulum where mass is concentrated at a point at the end of the (massless) string. Restoring force is the tangen3al force Ft = − mg sin θ d2s d2s ⇒ m 2 = − mg sin θ ⇒ 2 = − g sin θ dt dt For small ­angle oscilla3ons: Thus sin θ ≈ θ = d 2θ L 2 = − gθ dt d 2θ g ⇒ 2 = − θ = −ω 2θ dt L s L Therefore, ω= g L ⇒ T = 2π L g Comparing simple harmonic mo3on with uniform circular mo3on Uniform circular mo3on: Simple harmonic mo3on: θ (t ) = ω t + θ i x(t ) = R cosθ (t ) = R cos(ω t + θ i ) x (t ) = A cos (ω t + ϕ ) which is the same as for uniform circular mo3on if A = R and ϕ = θi. R θ x(t) Simple harmonic mo6on is the projec6on of uniform circular mo6on on a straight line. Projec3on Damped oscilla3ons What if we include fric3on or drag? In cases of air or liquid drag (as shown in the ﬁgure), there will be a drag force which is propor3onal to the velocity (but opposed to it) Drag force: Newton’s second law (ignore gravita3onal force or think of a horizontal set up of the spring for simplicity): Write as diﬀeren3al equa3on: The solu3on is: x(t ) = Ae − bt / 2 m cos (ω t + ϕ ) with ω= k b2 ⎛b⎞ − = ω2 − ⎜ 0 ⎝ 2m ⎟ ⎠ m 4 m2 2 where ω0 is the angular frequency of undamped harmonic oscillator: ω 0 = As we can see the amplitude of the oscilla3on is decreasing over 3me. k m Damped oscilla3ons (cotd…) Thus mechanical energy, E, is converted to heat via the fric3on or drag and is decreasing over 3me as well. where v(t ) = d d x(t ) = [ Ae− bt / 2 m cos(ω ' t + ϕ )] dt dt We see that the general expression for E(t) is rela3vely complicated. We can simplify it for the case of weak damping, i.e., for small b/m: 1 2 − bt / 2 m E (t ) ≈ 2 kA e Example: Consider three damped oscillators: All have m = 250 g and k =85 N/m, but oscillator (1) has b1 = 0, (2) has b2 = .070 kg/s and (3) has b3 = 5kg/s. What are (a) the periods of these oscillators and (b) how long does it take in each case unIl the oscillaIon amplitude has dropped to half of its iniIal value ? b12 k k 85N/m ω1 = − = = ω0 = = 18.44rad/s 2 m 4m m 0.25kg 2π ⇒ T1 = = 0.340s ω1 ω2 = 2 b2 k 85N/m (0.07 kg/s)2 − = − = 18.44rad/s ≈ ω 0 m 4 m2 0.25kg 4 ⋅ (0.25kg)2 (b) Condi3on: xa (t1/ 2 ) = xa (0 ) ⇒ Ae− bt 1 2 1/ 2 /2m = 1 A 2 where xa is the amplitude of the oscilla3on at some 3me t. ln e− bt1/2 / 2 m = ln ( ) 1 2 b32 k 85N/m (5kg/s)2 ω3 = − = − = 15.49rad/s m 4 m2 0.25kg 4 ⋅ 0.25kg 2π ⇒ T3 = = 0.406s ω3 (1)  t1/2 = ∞ (2) t1/2 = 4.95s (3) t1/2 = 0.07s Forced oscilla3ons If you are periodically applying a force on an oscillator (as in pushing somebody on a swing) you are ‘forcing’ the oscilla3on. If the force applied is harmonic and there is also some damping, the diﬀeren3al equa3on would look like this ( just add the periodic external to the net force in Newton’s second law): d 2x dx m 2 + b + kx = Fext sin ω t + ϕ ⇒ x = A cos(ω t + φ ) dt dt Fext / m A= 2 ⎛ bω ⎞ 2 (ω 2 − ω 0 )2 + ⎜ ⎝ m⎟ ⎠ ( ) But one thing is interes3ng: We have two angular frequencies: the ‘natural’ frequency of the oscillator given by ω 0 = k m and the frequency of the external force, ω, where / ω= k / m − b2 / 4 m 2 What happens when these two frequencies are (1) not the same or (2) the same ? Resonance A= Fext / m ⎛ bω ⎞ 2 (ω 2 − ω 0 )2 + ⎜ ⎝ m⎟ ⎠ 2 When frequency of the driving force ω approaches the natural frequency of the oscillator ω0 , the amplitude of the oscilla3on gets very large. This drama3c increase of amplitude is called resonance and the frequency at this occurs (ω = ω0) is called the resonant frequency. THE POWER OF RESONANCE CAN DESTROY A BRIDGE. ON NOVEMBER 7, 1940, THE ACCLAIMED TACOMA NARROWS BRIDGE COLLAPSED DUE TO OVERWHELMING RESONANCE ...
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