PY241 Solutions Set 9 (Dated: November 7, 2002)

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1 PY241 Solutions Set 9 (Dated: Noveber 7, 2002) 9-9 At what displaceent of an object undergoing siple haronic otion is the agnitude greatest for the... (a) velocity? The velocity is greatest at x = 0, the equilibriu position. At any other position, there is both inetic energy and spring energy, but at the equilibriu, all of the energy is inetic. Maxiu inetic energy axiu velocity (b) acceleration? Newton s law applied to a spring tells us: F = a = x a = x, thus acceleration is greatest when the spring is copressed/stretched the greatest. For an oscillation with aplitude A, the greatest acceleration is at x = ±A 9-15 When a 30 N ass is applied to a spring, it stretches 0.2. (a) If a 5 g ass is hung fro the spring and reains at rest, how uch is the string stretched fro its original length? Knowing that the spring stretches 0.2 with 30 N applied to it, we can solve for the spring constant. F = x 30 N = (0.2 ) = 150 N 1 (1) Now that we have, we can solve for the distance any force will stretch the spring. For a 5 g ass hanging on the spring, the resulting stretching is: g = x (5 g)(9.8 s 2 ) = (150 N 1 )(x) x =.326 (2) (b) What is the period of oscillation of the ass and the spring? The period is defined as: T = 1 ( 1 f = 2π T = (2π) ) 1 = (2π) 5 g = 1.15 s (3) 150 N A 50 g boy rides on a Pogo stic, a pole with a spring on its botto. lands on the ground, copressing the spring.05. He jups into the air 0.3 and (a) How uch energy is stored in the spring? By the conservation of energy, all of the gravitational potential energy at the pea of the jup ust be converted to spring potential energy when the boy hits the ground. U spring = GP E = gh = (50 g)(9.8 s 2 )(0.3 ) = 147 J (4)

2 2 (b) What is the spring constant? U spring = 1 2 x2 = 147 J 1 2 (0.05 )2 = 147 J = (2)(147 J) (0.05 ) 2 = 117, 600 N 1 (5) (c) What is the characteristic frequency of oscillation? 2π = 1 117, 600 N 1 = 7.72 Hz (6) 2π 50 g 9-42 The otolith in a fish has a ass of.022 g = g, and the effective spring constant is 3 N 1. (a) What is the characteristic frequency of the otolith? We can calculate the frequency using the sae procedure as before: 2π = 1 3 N 1 2π = Hz (7) g This oscillation is a daped. This eans that over tie the aplitude will decrease, as will the frequency. Therefore this frequency will only be correct very shortly after the onset of oscillation. (b) Is the frequency consistent with the idea that the otolith should respond rapidly to changes in orientation? The period corresponding to this frequency is T = 1/f =.017 s. This is a very short aount of tie, which agrees with the idea that the otolith responds rapidly to changes in orientation The huan leg can be approxiated by a cylinder. (a) Estiate the characteristic frequency of your legs, when swung fro the hip with the nee loced. In this proble we are treating the huan leg as a physical pendulu. physical pendulu is: gd 2π I The frequency associated with a (8) Where I is the oent of inertia for the object that is oscillating. If we approxiate waling otion as a rod (the leg) rotating about its end, we can then directly apply the forula above. The oent of inertia of a rod or cylinder rotating about its end it I = 1 3 l2. d is the distance fro the rotation axis to the center of gravity of the object. For

3 3 siplicity let us tae d = l/2. Plugging in to (8) gd 2π I g l 2 2π 1 3 l2 3g 2π 2l For the length of a leg, lets use l = 1. The frequency can now be calculated: 3(9.8 s 2 ) =.61 Hz (10) 2π 2(1.25 ) (9) (b) If noral waling were perfored with the legs swinging at their natural frequency, how far could you wal in 1 hour? The period corresponding to the frequency calculated above is T = 1/.61 = 1.64 s. This is how long it taes to coplete one stride. In order to calculate the distance travelled in an hour we need the distance covered in a single stride. Waling brisly, each stride is 2. Therefore: ( 1 stride )( 2 )( 3600 s ) = 4400 = 4.4 (11) 1.64 s 1 stride 1 hr 9-69 The siple haronic oscillator provides a good odel for sall oscillations about equilibriu in any systes, including olecules. In the H 2 olecule, the two hydrogen atos can oscillate toward and away fro each other, so that the center of ass reains stationary. Each oves as though connected to a spring with constant 1130 N 1. (a) Find the frequency of oscillation. Using H = g: 2π = 1 2π 1130 N g = Hz (12) (b) If the vibrational energy of the olecule is 1.23 ev, what is the aplitude of the oscillations for each ato? Each Hydrogen ato has half of the energy of the olecule. aplitude, all of the energy of the ato is stored in the spring : When the oscillations are at their axiu E ato E ato = P E spring =.5(1.23 ev ) =.615 ev = 1 2 A2.615 ev = J = 1 2 (1130 N 1 )A 2 A = Notice that we had to convert to Joules so that the units would end up correct 2( J) 1130 N 1 = (13)

4 4 (c)find the axiu velocity of the atos relative to the center of ass. Given an equation for SHM x(t) = A cos(ωt), the velocity as a function of tie is the tie derivative of the position. v(t) = ωa sin(ωt). The axiu value that v(t) can reach is v ax = ωa N v ax = ωa = g ( ) = s 1 (14) 9-73 Show that x(t) = A cos(ωt + φ) is a solution of the SHM equation a = x. a(t) = d2 x(t) d 2 t, so to test our function we ust tae two tie derivatives: d 2 d (A cos(ωt + φ)) = ωa sin(ωt + φ) dt dt 2 (A cos(ωt + φ)) = d dt ( ωa sin(ωt + φ)) = ω2 A cos(ωt + φ) = ω 2 x(t) (15) By definition, ω = ω2 = a(t) = ω 2 x(t) = x(t) (16) Our choice of x(t) = A cos(ωt + φ) as a solution to the SHM equation is correct. (b) What is the significance of φ. φ is a constant, nown as the phase. It describes the starting point of the oscillation. The equation x(t) = A cos(ωt) only describes an oscillation has x(0) = A. The introduction of φ allows us to describe an oscillation with any starting position Since various parts of the body have characteristic frequencies of vibration, we can thin of these parts as being connected by springs. The spring is fored by the flexible connections of these parts of the body. In Ch. 8 we learned that the spring constant is proportional to the cross sectional area divided by the length of the spring aterial. (a) Using scaling hypothesis of 8.6 l r 2/3, show that the spring constant 1/2. Fro Ch. 8 we now that = EA l. Using A r 2, and l r 2/3 r l 3/2 we can see how scales: r2 l (l 3/2 ) 2 l l 2 (17) The scaling law in 8.6 lead to the conclusion that l 4 l 1/4. Therefore: l 2 1/2 (18) (b) Show that the characteristic frequency should scale as f 1/4. f 1/2 1/2 1/4 (19)

5 The abdoen and thorax of a 60 g huan has a resonance at about 3 Hz. (a) Using the result of proble 9-82, what would you expect the corresponding characteristic frequency to be in a 20 g = 2 2 g ouse? Fro the previous proble, we now that f 1/4. We can write this relationship as = c 1/4, where c is a constant. We can solve for c using the nuerical values for the resonant frequency of a huan abdoen: f = c 1/4 c = f = 3 Hz 1/4 (60 g) = /4 g1/4 s (20) Applying the relationship to the ouse, with the correct value of c gives us the guess for the resonant frequency of a ouse abdoen: f ouse = c 1/4 ouse = (8.35 g 1/4 s)( g) 1/4 = 22 Hz (21) (b) Experientally the frequency in ice is between 18 and 25 Hz. How does this copare to the result in part (a)? The calculated value of 22 Hz is within the easured range. body parts is correct in this experient. The scaling law for resonant frequencies of

Periodic Motion is everywhere

Periodic Motion is everywhere Lecture 19 Goals: Chapter 14 Interrelate the physics and atheatics of oscillations. Draw and interpret oscillatory graphs. Learn the concepts of phase and phase constant. Understand and use energy conservation