Conceptual Q: 4 (7), 7 (), 8 (6) Physics 4 HW Set Chapter 5 Serway 8 th ( 7 th ) Q4(7) Answer (c). The equilibrium position is 5 cm below the starting point. The motion is symmetric about the equilibrium position, so the two turning points are 30 cm apart. Q 7() Use: du F dx (a) No force is exerted on the particle. The particle moves with constant velocity. The particle feels a constant force toward the left. It moves with constant acceleration toward the left. If its initial push is toward the right, it will slow down, turn around, and speed up in motion toward the left. If its initial push is toward the left, it will just speed up. (c) A constant force towards the right acts on the particle to produce constant acceleration toward the right. (d) The particle moves in simple harmonic motion about the lowest point of the potential energy curve. Q5.6 Yes. An oscillator with damping can vibrate at resonance with amplitude that remains constant in time. Without damping, the amplitude would increase without limit at resonance. Problems: 4 (), 5 (3), 8 (6), 8 (4), 0 (8), 9 (9), 3 (5), 5 (47), 64 & 67 P4 () In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression x = (5.00 cm) cos(t + /6) where x is in centimeters and t is in seconds. At t = 0, find (a) the position of the piston, its velocity, and (c) its acceleration. (d) Find the period and amplitude of the motion. (a) x 5.00 cm cos t 6 t, x At 0 5.00 cm cos 4.33 cm 6 dx 0.0 cm s sin t dt 6 v dv (c) a (d) 0.0 cm s cos t dt 6 A 5.00 cm and At t 0, v 5.00 cm s At t 0, a 7.3 cm s T 3.4 s
P 5(3) The position of a particle is given by the expression x = (4.00 m) cos(3.00 t + ), where x is in meters and t is in seconds. Determine (a) the frequency and period of the motion, the amplitude of the motion, (c) the phase constant, and (d) the position of the particle at t = 0.50 s. x 4.00 m cos 3.00 t Compare this with x A cos t to find (a) f 3.00 or f.50 H z T f 0.667 s (c) A 4.00 m rad (d) xt 0.50 s 4.00 m cos.75.83 m P8 (6) A simple harmonic oscillator takes.0 s to undergo five complete vibrations. Find (a) the period of its motion, the frequency in hertz, and (c) the angular frequency in radians per second. (a).0 s T 5.40 s f T.40 0.47 Hz (c) f 0.47.6 rad s P8 (4) A 00-g block is attached to a horizontal spring and executes simple harmonic motion with a period of 0.50 s. If the total energy of the system is.00 J, find (a) the force constant of the spring and the amplitude of the motion. m 00 g, T 0.50 s, E.00 J; 5. rad s T 0.50 k m 0.00 kg 5. rad s 6 N m (a) ka E E A.00 k 6 0.78 m
P0(8) A.00-kg object is attached to a spring and placed on a horizontal, smooth surface. A horizontal force of 0.0 N is required to hold the object at rest when it is pulled 0.00 m from its equilibrium position (the origin of the x axis). The object is now released from rest with an initial position of xi = 0.00 m, and it subsequently undergoes simple harmonic oscillations. Find (a) the force constant of the spring, the frequency of the oscillations, and (c) the maximum speed of the object. Where does this maximum speed occur? (d) Find the maximum acceleration of the object. Where does it occur? (e) Find the total energy of the oscillating system. Find (f) the speed and (g) the acceleration of the object when its position is equal to one third of the maximum value. F 0.0 N (a) k 00 N m x 0.00 m k 50.0 rad s so f.3 H z m (c) v A m ax 50.0 0.00.4 m s at x 0 (d) a A m ax 50.0 0.00 0.0 m s at x A 00 0.00.00 J (e) E ka 8 v A x 50.0 0.00.33 m s 9 (f) (g) 0.00 a x 50.0 3.33 m s 3 P9 A physical pendulum in the form of a planar body moves in simple harmonic motion with a frequency of 0.450 Hz. If the pendulum has a mass of.0 kg and the pivot is located 0.350 m from the center of mass, determine the moment of inertia of the pendulum about the pivot point. T ; f f 0.450 H z, d 0.350 m, and m.0 kg I T ; m gd T 4 I m gd m gd m gd.0 9.80 0.350 0.944 kg m I T 4 f 4 4 0.450 s FIG. P5.35
P3(5) A simple pendulum has a mass of 0.50 kg and a length of.00 m. It is displaced through an angle of 5.0 and then released. What are (a) the maximum speed, the maximum angular acceleration, and (c) the maximum restoring force? What If? Solve this problem by using the simple harmonic motion model for the motion of the pendulum, and then solve the problem more precisely by using more general principles. Using the simple harmonic motion model: A r m 5 0.6 m 80 g 9.8 m s 3.3 rad s L m FIG. P5.3 (a) vm ax A 0.6 m 3.3 s 0.80 m s am ax A 0.6 m 3.3 s.57 m s (c) atan r atan.57 m s r m F m a 0.5 kg.57 m s 0.64 N.57 rad s More precisely, (a) m gh m v and hl cos vm ax gl cos 0.87 m s I m gl sin m gl sin g sin i ml L m ax.54 rad s Fm ax m gsin i 0.50 9.80 sin5.0 0.634 N (c)
P 5 (47) A small ball of mass M is attached to the end of a uniform rod of equal mass M and length L that is pivoted at the top (Fig. P5.5). (a) Determine the tensions in the rod at the pivot and at the point P when the system is stationary. Calculate the period of oscillation for small displacements from equilibrium, and determine this period for L =.00 m. (Suggestions: Model the object at the end of the rod as a particle and use Eq. 5.8.) Let F represent the tension in the rod. (a) At the pivot, F M g M g M g y A fraction of the rod s weight Mg L as well as the weight of the ball pulls down on point P. Thus, the tension in the rod at point P is y y F M g M g M g L L. Relative to the pivot, 4 I Irod Iball M L M L M L 3 3 pivot P L y M FIG. P5.5 I For the physical pendulum, T where m M and d is the distance from the pivot m gd to the center of mass of the rod and ball combination. Therefore, For L.00 m, L M M L 4 3L 3 ML 4 L d and T. M M 4 3L M g 4 3 g 4.00 m T.68 s. 3 9.80 m s
P.67 A block of mass m is connected to two springs of force constants k and k as shown In each case, the block moves on a frictionless table after it is displaced from equilibrium and released. Show that in the two cases the block exhibits simple harmonic motion with periods (a) T m k k k k T m k k When the mass is displaced a distance x from equilibrium, spring is stretched a distance x and spring is stretched a distance x. By Newton s third law, we expect kx kx. When this is combined with the requirement that x x x, k we find x k k kk The force on either spring is given by F x ma k k where a is the acceleration of the mass m. x This is in the form and F keff x ma m m k k T k k k eff In this case each spring is distorted by the distance x which the mass is displaced. Therefore, the restoring force is F k k x and keff k k so that T m k k
T^ (s^) 64. When a block of mass M, connected to the end of a spring of mass ms = 7.40 g and force constant k, is set into simple harmonic motion, the period of its motion is () T M m s / 3 k A two-part experiment is conducted with the use of blocks of various masses suspended vertically from the spring, as shown. (a) Static extensions of 7.0, 9.3, 35.3, 4.3, 47., and 49.3 cm are measured for M values of 0.0, 40.0, 50.0, 60.0, 70.0, and 80.0 g, respectively. Construct a graph of Mg versus x, and perform a linear least-squares fit to the data. From the slope of your graph, determine a value for k for this spring. The system is now set into simple harmonic motion, and periods are measured with a stopwatch. With M = 80.0 g, the total time for 0 oscillations is measured to be 3.4 s. The experiment is repeated with M values of 70.0, 60.0, 50.0, 40.0, and 0.0 g, with corresponding times for 0 oscillations of.5,.67, 0.67, 9.6, and 7.03 s. Compute the experimental value for T from each of these measurements. Plot a graph of T versus M and determine a value for k from the slope of the linear least-squares fit through the data points. Compare this value of k with that obtained in part (a). (c) Obtain a value for ms from your graph and compare it with the given value of 7.40 g. (a) m x F 0 0.7 0.96 40 0.93 0.39 50 0.353 0.49 60 0.43 0.588 70 0.47 0.686 80 0.493 0.784 From the plot to 3 sig figs, 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0. 0. 0 0 0. 0.4 0.6 y =.7386x - 0.8 R = 0.9884 Series Linear (Series) k.74 N / m 6% Since 0.3/.74x00 = 6.4% With M = 80 g and t = 3.4s for 0 oscillations, the period is.34s for one oscillation. T (s) M(kg) T^ (s^) 0.703 0.0 0.494 0.96 0.04 0.95.067 0.05.38.67 0.06.36.5 0.07.568.34 0.08.798.000.500.000 0.500 T^ vs Mass y =.665x + 0.0589 R = 0.9999 Series Linear (Series) 0.000 0 0.0 0.04 0.06 0.08 0. Mass (kg)
4 4 4 Rewrite equation (): T M ms T ( M ms) k 3k k 3 and set it equal to the equations from the graph: T =.665M + 0.058 and solve for k, with M =.0.08kg, and the mass of the spring is 7.40g =.0074kg M ms.08.0074 4 3 3.08467 T ( M ms ).7M 0.0589 k 4 4 4 k 3.7M 0.0589.7(.08) 0.0589.7949 k.8384 0.0589/.8384*00 3.5% So from the plot and using equation (): k.8 N / m 3% Comparing to the results of part (a) k.74 N / m 6% The discrepancy and percent differency is: 0.07 N/ m 4% c) To find the mass of the spring set M =0: 4 4 T ( M ms).7m 0.0589 ms 0.0589 k 3.8 3 ms 8.g With a discrepancy and % difference from 7.40g of 0.7g 9%