Simple Harmonic Motion

Transcription

1 SHM-1 Siple Haronic Motion A pendulu, a ass on a spring, and any other kinds of oscillators ehibit a special kind of oscillatory otion called Siple Haronic Motion (SHM). SHM occurs whenever : i. there is a restoring force proportional to the displaceent fro equilibriu: F ii. the potential energy is proportional to the square of the displaceent: PE iii. the period T or frequency f = 1 / T is independent of the aplitude of the otion. iv. the position, the velocity v, and the acceleration a are all sinusoidal in tie. v t t (Sinusoidal eans sine, cosine, or anything in between.) As we will see, any one of these four properties guarantees the other three. If one of these 4 things is true, then the oscillator is a siple haronic oscillator and all 4 things ust be true. Not every kind of oscillation is SHM. For instance, a perfectly elastic ball bouncing up and down on a floor: the ball's position (height) is oscillating up and down, but none of the 4 conditions above is satisfied, so this is not an eaple of SHM. A ass on a spring is the siplest kind of Siple Haronic Oscillator. k F restore F restore relaed: = 0 Hooke's Law: F spring = k ( ) sign because direction of F spring is opposite to the direction of displaceent vector (bold font indicates vector) k = spring constant = stiffness, units [k] = N / Big k = stiff spring Definition: aplitude A = a = in. Mass oscillates between etree positions = +A and = A A +A 0 4/10/009 University of Colorado at Boulder

2 SHM- Notice that Hooke's Law (F = k) is condition i : restoring force proportional to the displaceent fro equilibriu. We showed previously (Work and Energy Chapter) that for a spring obeying Hooke's Law, the potential energy is U = (1/)k, which is condition ii. Also, in the chapter on Conservation of Energy, we showed that F = du/d, fro which it follows that condition ii iplies condition i. Thus, Hooke's Law and quadratic PE (U ) are equivalent. We now show that Hooke's Law guarantees conditions iii (period independent of aplitude) and iv (sinusoidal otion). We begin by deriving the differential equation for SHM. A differential equation is siply an equation containing a derivative. Since the otion is 1D, we can drop the vector arrows and use sign to indicate direction. F = a and F = k a = k net net d k a = dv/ = d / = The constants k and and both positive, so the k/ is always positive, always. For notational convenience, we write k/ = ω. (The square on the ω reinds us that ω is always positive.) The differential equation becoes d = ω (equation of SHM) This is the differential equation for SHM. We seek a solution = (t) to this equation, a function = (t) whose second tie derivative is the function (t) ultiplied by a negative constant ( ω = k/). The way you solve differential equations is the sae way you solve integrals: you guess the solution and then check that the solution works. Based on observation, we guess a sinusoidal solution: (t) = A cos( ω t +ϕ ), where A, ϕ are any constants and (as we'll show) A = aplitude: oscillates between +A and A ϕ = phase constant (ore on this later) k ω=. Danger: ωt and ϕ have units of radians (not degrees). So set your calculators to radians when using this forula. Just as with circular otion, the angular frequency ω for SHM is related to the period by π ω= π f =, T = period. T 4/10/009 University of Colorado at Boulder

3 SHM-3 (What does SHM have to do with circular otion? We'll see later.) Let's check that (t) = Acos( ω t + ϕ) is a solution of the SHM equation. d = = ω ω +ϕ. d dcos( ) d cos( ω t +ϕ ) =, ( =ω t +ϕ) d d t Taking the first derivative d/, we get v(t) A sin ( t ) Here, we've used the Chain ule: Taking a second derivative, we get [ Acos( t )] = sin ω = ωsin( ω t +ϕ) d dv d a(t) = = = ( Aωsin( ω t+ϕ )) = Aω cos( ω t + ϕ) d d = ω ω +ϕ = ω This is the SHM equation, with k k ω =, ω = We have shown that our assued solution is indeed a solution of the SHM equation. (I leave to the atheaticians to show that this solution is unique. Physicists seldo worry about that kind of thing, since we know that nature usually provides only one solution for physical systes, such as asses on springs.) We have also shown condition iv:, v, and a are all sinusoidal functions of tie: ( ) (t) = Acos ω t+ϕ v(t) = Aω sin( ω t +ϕ) a(t) = A ω cos( ω t +ϕ) k π The period T is given by ω= = T = π. We see that T does not T k depend on the aplitude A (condition iii). Let's first try to ake sense of ω= k/: big ω eans sall T which eans rapid oscillations. According to the forula, we get a big ω when k is big and is sall. This akes sense: a big k (stiff spring) and a sall ass will indeed produce very rapid oscillations and a big ω. 4/10/009 University of Colorado at Boulder

4 SHM-4 A closer look at (t) = A cos(ωt+ϕ) Let's review the sine and cosine functions and their relation to the unit circle. We often define the sine and cosine functions this way: hypotenuse adjacent opposite adj cos = hyp opp sin = hyp This way of defining sine and cosine is correct but incoplete. It is hard to see fro this definition how to get the sine or cosine of an angle greater than 90 o. A ore coplete way of defining sine and cosine, a way that gives the value of the sine and cosine for any angle, is this: Draw a unit circle (a circle of radius r = 1) centered on the origin of the -y aes as shown: Define sine and cosine as adj cos = = = hyp 1 opp y sin = = = y hyp 1 r = 1 y point (, y) This way of defining sin and cos allows us to copute the sin or cos of any angle at all. For instance, suppose the angle is = 10 o. Then the diagra looks like this: The point on the unit circle is in the third quadrant, where both and y are negative. So both cos = and sin = y are negative y 1 point (, y) For any angle, even angles bigger than 360 o (ore than once around the circle), we can always copute sin and cos. When we plot sin and cos vs angle, we get functions that oscillate between +1 and 1 like so: 4/10/009 University of Colorado at Boulder

5 SHM cos = = π rad +1 1 sin = π We will alost always easure angle in radians. Once around the circle is π radians, so sine and cosine functions are periodic and repeat every tie increases by π rad. The sine and cosine functions have eactly the sae shape, ecept that sin is shifted to the right copared to cos by = π/. Both these functions are called sinusoidal functions. +1 = π/ 1 cos sin The function cos( + ϕ) can be ade to be anything in between cos() and sin() by adjusting the size of the phase ϕ between 0 and π. π cos, ( ϕ= 0) sin = cos, ϕ= π/ ( ) The function cos(ωt + ϕ) oscillates between +1 and 1, so the function Acos(ωt + ϕ) oscillates between +A and A. Acos (ωt) =π +A A =ωt π Why ω=? The function f() = cos is periodic with period = π. Since T =ωt+ϕ, and ϕ is soe constant, we have =ω t. One coplete cycle of the cosine function corresponds to = π and t = T, (T is the period). So we have π = ω T or π t ω=. Here is another way to see it: cos( ω t) = cos π is periodic with period t T T 4/10/009 University of Colorado at Boulder

6 SHM-6 = T. To see this, notice that when t increases by T, the fraction t/t increases by 1 and the fraction πt/t increases by π. Acos (ωt) Acos (ωt) +A +A t ωt A A t = T (ωt) = π Now back to siple haronic otion. Instead of a circle of radius 1, we have a circle of radius A (where A is the aplitude of the Siple Haronic Motion). SHM and Conservation of Energy: ecall PE elastic = (1/) k = work done to copress or stretch a spring by distance. If there is no friction, then the total energy E tot = KE + PE = constant during oscillation. The value of E tot depends on initial conditions where the ass is and how fast it is oving initially. But once the ass is set in otion, E tot stays constant (assuing no dissipation.) At any position, speed v is such that v + k = E. 1 1 tot When = A, then v = 0, and all the energy is PE: KE + PE = E So total energy E = ka 1 tot 0 (1/ )ka When = 0, v = v a, and all the energy is KE: KE + PE = E So, total energy E tot = v. 1 a (1/ )va 0 tot tot 4/10/009 University of Colorado at Boulder

7 SHM-7 y (1/)k E tot = KE KE + PE PE A +A So, we can relate v a to aplitude A : PE a = KE a = E tot ka = v 1 1 a v a = k A Eaple Proble: A ass on a spring with spring constant k is oscillating with aplitude A. Derive a general forula for the speed v of the ass when its position is. Answer: k v() = A 1 A Be sure you understand these things: range of otion = A v = 0 PE = a KE = in F = a a = a = 0 v = a PE = in KE = a F = 0 a = 0 4/10/009 University of Colorado at Boulder

8 SHM-8 Pendulu Motion A siple pendulu consists of a sall ass suspended at the end of a assless string of length L. A pendulu eecutes SHM, if the aplitude is not too large. L Forces on ass : y F T = tension g sin g cos = / L (rads) g The restoring force is the coponent of the force along the direction of otion: restoring force = g sin g = g L h Clai: sin (rads) when is sall. sin = L s L = s h If sall, then h s, and L, so sin. Try it on your calculator: = 5 o = rad, sin = g Frestore = L is eactly like Hooke's Law F restore = k, ecept we have replaced the constant k with another constant (g / L). The ath is eactly the sae as with a ass on a spring; all results are the sae, ecept we replace k with (g/l). T T spring = π pend k = π = π L g ( g/l) Notice that the period is independent of the aplitude; the period depends only on length L and acceleration of gravity. (But this is true only if is not too large.) 4/10/009 University of Colorado at Boulder

9 SHM-9 Appendi: SHM and circular otion There is an eact analogy between SHM and circular otion. Consider a particle oving with constant speed v around the ri of a circle of radius A. The -coponent of the position of the particle has eactly the sae atheatical for as the otion of a ass on a spring eecuting SHM with aplitude A. A v angular velocity = ω t so d ω= = const A +A = A cos = A cos ω t This sae forula also describes the sinusoidal otion of a ass on a spring. That the sae forula applies for two different situations (ass on a spring & circular otion) is no accident. The two situations have the sae solution because they both obey the sae equation. As Feynan said, "The sae equations have the sae d solutions". The equation of SHM is = ω. We now show that a particle in circular otion obeys this sae SHM equation. ecall that for circular otion with angular speed ω, the acceleration of a the particle is v toward the center and has agnitude a =. Since v = ω, we can rewrite this as a ( ω ) = = ω Let's set the origin at the center of the circle so the position vector is along the radius. Notice that the acceleration vector a is always in the direction opposite the position vector. Since a = ω, the vectors a and are related by a = ω. The -coponent of this vector equation is: d 0 a = ω. If we write =, then we have = ω, which is the SHM equation. Done. v a A 0 ωt +A 4/10/009 University of Colorado at Boulder