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Download Ch 14 HW Day 2 p 455 – 464
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Ch 14 HW Day 2 p 455 – 464, #’s 33-39 (ODDS), 40*, 41-59* (ODDS, omit #53) Simple Harmonic Motion and Circular Motion 33 • Picture the Problem We can find the period of the motion from the time required for the particle to travel completely around the circle. The frequency of the motion is the reciprocal of its period and the x-component of the particle’s position is given by x A cost . (b) Use the definition of speed to find the period of the motion: (a) Because the frequency and the period are reciprocals of each other: (c) Express the x component of the position of the particle: Assuming that the particle is on the positive x axis at time t = 0: Substitute for A, , and to obtain: T = 2/; Or, use 2r/v = T, subst, 2(0.4m)/.8m/s s Subst will give you 2/s 2s-1 (use for ) T 2r 2 0.4 m 3.14 s v 0.8 m/s f 1 1 0.318 Hz T 3.14 s x A cost A Acos cos 1 1 0 x A cos2ft 40 cm cos2 s1 t Energy in Simple Harmonic Motion 35 • Picture the Problem The total energy of the object is given by Etot 12 kA2 , where A is the amplitude of the object’s motion. Express the total energy of the system: Etot 12 kA2 Substitute numerical values and evaluate Etot: Etot 1 2 4.5 kN/m0.1m2 22.5 J 37 • Picture the Problem The total mechanical energy of the oscillating object can be expressed in terms of its kinetic energy 2 1 as it passes through its equilibrium position: Etot 2 mvmax . Its total energy is also given by Etot 2 kA . (E is conserved, so when v is a max, K is a max and U is a min, or 0.) We can equate these expressions to obtain an expression for A. 1 2 (a) Express the total mechanical energy of the object in terms of its maximum kinetic energy: 2 E 12 mvmax Substitute numerical values and evaluate E: E (b) Express the total energy of the object in terms of the amplitude of its motion: Etot 12 kA2 1 2 1.5 kg0.7 m/s 2 0.368 J Solve for A: Substitute numerical values and evaluate A: A 2 Etot k A 20.368 J 3.84 cm 500 N/m 39 • Picture the Problem The total energy of the object is given by Etot 12 kA2 . We can solve this equation for the force constant k and substitute the numerical data to determine its value. Express the total energy of the oscillator as a function of the amplitude of its motion: Etot 12 kA2 Solve for k: k 2 E tot A2 Substitute numerical values and evaluate k: k 21.4 J 1.38 kN/m 2 0.045 m *40 •• Picture the Problem The total energy of the object is given, in 2 . We can terms of its maximum kinetic energy by Etot 12 mvmax express vmax in terms of A and and, in turn, express in terms of amax to obtain an expression for Etot in terms of amax. Express the total energy of the object in terms of its maximum kinetic energy: 2 Etot 12 mvmax Relate the maximum speed of the object to its angular frequency: Substitute to obtain: Relate the maximum acceleration of the object to its angular frequency: Substitute and simplify to obtain: Substitute numerical values and evaluate Etot: vmax A (v is the first derivative of x; sin and cos have max magnitudes at 1, so those portions of the respective eqs 1, vmax = A∙1, or just A. This generalization is true, respectively, for amax, as well, so amax, which = d2x/dt2, = A2.) Etot 12 m A 12 mA2 2 2 amax A 2 or 2 amax A Etot 12 mA2 amax 1 2 mAamax A Etot 12 3 kg 0.08 m 3.50 m/s 2 0.420 J Springs 41 • Picture the Problem The frequency of the object’s motion is given by f 1 2 k m. (f = 1/2 ω and = k m ). Its period is the reciprocal of its frequency. The maximum velocity and acceleration of an object executing simple harmonic motion are vmax A and amax A 2 , respectively. (a) The frequency of the motion is given by: f 1 2 k m Substitute numerical values and evaluate f: f 1 2 4.5 kN/m 6.89 Hz 2.4 kg Remember to put 2 in parentheses, or you’ll get 68.0 as your answer! 1 1 0.145 f 6.89 s 1 (b) The period of the motion to is the reciprocal of its frequency: (c) Because the object is released from rest after the spring to which it is attached is stretched 10 cm: T (d) Express the object’s maximum speed: vmax A 2fA Substitute numerical values and evaluate vmax: vmax 2 6.89 s 1 0.1m 4.33 m/s (e) Express the object’s maximum acceleration: amax A 2 vmax 2fvmax Substitute numerical values and evaluate amax: amax 2 6.89 s 1 4.33 m/s (f) The object first reaches its equilibrium when: t 14 T 14 0.145 s 0.0363 s A 0.100 m 187 m/s 2 See #41 scanned image Because the resultant force acting on the object as it passes through its equilibrium point is zero, the acceleration of the a 0 object is: 43 • Picture the Problem The angular frequency, in terms of the force constant of the spring and the mass of the oscillating object, is given by k m . The period of the motion is the reciprocal of its frequency. The maximum velocity and acceleration of an object executing simple harmonic motion are 2 2 vmax A and amax A , respectively. (a) Relate the angular frequency of the motion to the force constant of the spring: 2 k m or k m 2 4 2 f 2 m 3 kg Substitute numerical values to obtain: k 4 2 2.4 s 1 (b) Relate the period of the motion to its frequency: (c) Express the maximum speed of the object: T Substitute numerical values and evaluate vmax: vmax 2 2.4 s1 0.1m 1.51m/s (d) Express the maximum acceleration of the object: amax A 2 4 2 f 2 A Substitute numerical values and evaluate amax: amax 4 2 2.4 s 1 2 682 N/m 1 1 0.417 s f 2.4 s 1 vmax A 2fA 0.1m 2 22.7 m/s 2 45 • Picture the Problem We can relate the force constant k to the maximum acceleration by eliminating 2 between 2 k m and amax A 2 . We can also express the frequency f of the motion by substituting mamax/A for k in k m (a) Relate the angular frequency of the motion to the force constant and the mass of the oscillator: 2 Relate the object’s maximum acceleration to its angular frequency and amplitude and solve for the square of the angular frequency: Substitute to obtain: amax A 2 f 1 2 k . m or k 2m or 2 amax A (1) k Substitute numerical values and evaluate k: (b) Replace in equation (1) by 2f and solve for f to obtain: Substitute numerical values and evaluate f: k (c) The period of the motion is the reciprocal of its frequency: mamax A 4.5 kg 26 m/s 2 3.8 102 m 3.08 kN/m f 1 2 amax A f 1 2 26 m/s 2 4.16 Hz 3.8 102 m T 1 1 0.240 s f 4.16 s 1 47 •• Picture the Problem The maximum speed of the block is given by vmax A and the angular frequency of the motion is k m 5.48 rad/s . We’ll assume that the position of the block is given by x A cos t and solve for t for x = 4 cm and x = 0. We can use these values for t to find the time for the block to travel from x = 4 cm to its equilibrium position. (a) Express the maximum speed of the block as a function of the system’s angular frequency: Substitute numerical values and evaluate vmax: (b) Assuming that x A cos t, evaluate t for x = 4 cm = A/2: Evaluate v for t 3: vmax A vmax 0.08 m 5.48 rad/s 0.438 m/s A 1 A cos t t cos 1 2 2 3 I would write this as: 4 = 8 cos t. So, t = cos -1 (1/2). Then, they subst that into the equation for v. v vmax sin t 0.438 m/s sin 0.438 m/s 3 3 0.379 m/s 2 Express a as a function of vmax and : a A 2 cos t vmax cos t Substitute numerical values and evaluate a: a 0.438 m/s 5.48 rad/s cos 3 1.20 m/s 2 (c) Evaluate t for x = 0: 0 A cos t t cos 1 0 2 Let t = time to go from t 3 to t 2 . Then: Solve for and evaluate t: t t 2 3 6 95.5 ms 6 65.48 rad/s Or, you could find the entire period, and divide it by 4! 49 •• Picture the Problem Let the system include the object and the spring. Then, the net external force acting on the system is zero. Choose Ei = 0 and apply the conservation of mechanical energy to the system. 2 Express the period of the motion in terms of its angular frequency: T Apply conservation of energy to the system: Ei Ef or 0 U g U spring Substitute for Ug and Uspring: Solve for 2 = k/m: 0 mgx 12 k x Substitute numerical values and evaluate 2: Substitute in equation (1) to obtain: (1) 2 2 k 2g m x 2 29.81m/s 2 574 rad/s 2 2 3.42 10 m T 2π 574 rad/s 2 0.262 s 51 •• Picture the Problem The stunt woman’s kinetic energy, after 2 s of flight, is K2 s 12 mv22s . We can evaluate this quantity as soon as we know how fast she is moving after two seconds. Because her motion is oscillatory, her velocity as a function of time is vt A sin t . We can find the amplitude of her motion from her distance of fall and the angular frequency of her motion by applying conservation of energy to her fall to the ground. Express the kinetic energy of the stunt woman when she has fallen for 2 s: Express her velocity as a function of time: Letting Ei = 0, use conservation of energy to find the force constant of the elastic band: K 2s 12 mv22s (1) vt A sin t where = 0 (she starts from rest with positive displacement) and A 12 192 m 96 m (2) vt 96 m sin t 0 U g U elastic or 0 mgh 12 kh2 0 Solve for k: k 2mg h Substitute numerical values and evaluate k: k 260 kg 9.81m/s 2 6.13 N/m 192 m Express the angular frequency of her motion: Substitute numerical values and evaluate : k m 6.13 N/m 0.320 rad/s 60 kg Substitute in equation (2) to obtain: Evaluate v(2 s): vt 96 m0.320 rad/s sin 0.320 rad/s t 30.7 m/s sin 0.320 rad/s t v2 s 30.7 m/s sin 0.320 rad/s 2 s 18.3 m/s Substitute in equation (1) and evaluate K(2 s): K 2 s 1 2 60 kg18.3 m/s 2 10.1kJ 55 •• Picture the Problem The maximum height above the floor to which the object rises is the sum of its initial distance from the floor and the amplitude of its motion. We can find the amplitude of its motion by relating it to the object’s maximum speed. Because the object initially travels downward, it will be three-fourths of the way through its cycle when it first reaches its maximum height. We can find the minimum initial speed the object would need to be given in order for the spring to become uncompressed by applying conservation of energy. (a) Relate h, the maximum height above the floor to which the object rises, to the amplitude of its motion: h = A + 5.0 cm Relate the maximum speed of the object to the angular frequency and amplitude of its motion and solve for the amplitude: vmax A Using its definition, express and evaluate the force constant of the spring: Substitute numerical values in equation (2) and evaluate A: Substitute in equation (1) to obtain: (1) or A vmax k m k (2) mg 2 kg 9.81 m/s 2 654 N/m y 0.03 m A 0.3 m/s 2 kg 1.66 cm 654 N/m h 1.66 cm 5.00 cm 6.66 cm (b) Express the time required for the object to reach its maximum height the first time: t 34 T Express the period of the motion: T 2 m k Substitute numerical values and evaluate T: T 2 2 kg 0.347 s 654 N/m Substitute to obtain: t (c) Because h < 8.0 cm: Using conservation of energy and letting Ug be zero 5 cm above the floor, relate the height to which the object rises, y, to its initial kinetic energy: Because y L yi : 3 4 0.347 s 0.261s the spring is never uncompress ed. K U g U s 0 or, because Kf = Ui = 0, 2 2 1 1 2 mvi mgy 2 k y 2 12 k L yi 0 1 2 mvi2 mgy 12 k y 12 k y 0 2 2 and 1 2 Solve for and evaluate vi for y = 3 cm: mvi2 mgy 0 vi 2 gy 2 9.81 m/s 2 3 cm 0.767 m/s i.e., the minimum initial velocity that must be given to the object for the spring to be uncompressed at some time is 0.767 m/s Energy of an Object on a Vertical Spring 57 •• Picture the Problem Let the origin of our coordinate system be at y0, where y0 is the equilibrium position of the object and let Ug = 0 at this location. Because Fnet = 0 at equilibrium, the extension of the spring is then y0 = mg/k, and the potential energy stored in the spring is U s 12 ky02 . A further extension of the spring by an amount y increases Us to 2 2 2 2 2 1 1 1 1 1 2 k y y0 2 ky kyy0 2 ky0 2 ky mgy 2 ky0 . Consequently, if we set U = Ug + Us = 0, a further extension of the spring by y increases Us by ½ky2 + mgy while decreasing Ug by mgy. Therefore, if U = 0 at the equilibrium position, the change in U is given by 2 1 2 k y ' , where y = y y0. (a) Express the total energy of the system: E 12 kA2 Substitute numerical values and evaluate E: E (b) Express and evaluate Ug when the object is at its maximum downward displacement: (c) When the object is at its maximum downward displacement: U g mgA (d) The object has its maximum kinetic energy when it is passing through its equilibrium position: K max 12 kA2 1 2 600 N/m0.03 m2 0.270 J 2.5 kg 9.81m/s 2 0.03 m 0.736 J U s 12 kA2 mgA 1 2 600 N/m0.03 m2 2.5 kg 9.81m/s 2 0.03 m 1.01J 1 2 600 N/m0.03 m2 0.270 J *59 •• Picture the Problem We can find the amplitude of the motion by relating it to the maximum speed of the object. Let the origin of our coordinate system be at y0, where y0 is the equilibrium position of the object and let Ug = 0 at this location. Because Fnet = 0 at equilibrium, the extension of the spring is then y0 = mg/k, and the potential energy stored in the spring is U s 12 ky02 . A further extension of the spring by an amount y increases Us to 2 2 2 2 2 1 1 1 1 1 2 k y y0 2 ky kyy0 2 ky0 2 ky mgy 2 ky0 . Consequently, if we set U = Ug + Us = 0, a further extension of the spring by y increases Us by ½ky2 + mgy while decreasing Ug by mgy. Therefore, if U = 0 at the equilibrium position, the change in U is given by 12 k y'2 , where y = y y0. (a) Relate the maximum speed of the object to the amplitude of its motion: Solve for A: vmax A A vmax vmax Substitute numerical values and evaluate A: A 0.3 m/s (b) Express the energy of the object at maximum displacement: E 12 kA2 Substitute numerical values and evaluate E: E (c) At maximum displacement from equilibrium: U g mgA Substitute numerical values and evaluate Ug: (d) Express the 1 2 m k 1.2 kg 1.90 cm 300 N/m 300 N/m0.019 m2 0.0542 J U g 1.2 kg 9.81m/s 2 0.019 m 0.224 J U s 12 kA2 mgA potential energy in the spring when the object is at its maximum downward displacement: Substitute numerical values and evaluate Us: Us 300 N/m0.019 m2 1.2 kg 9.81m/s 2 0.019 m 1 2 0.278 J