# phys124s10-hw10.pdf

## Physics 124 with Gilman at Rutgers University - New Brunswick/Piscataway *

- StudyBlue
- New-jersey
- Rutgers University - New Brunswick/Piscataway
- Physics
- Physics 124
- Gilman
- phys124s10-hw10.pdf

Advertisement

HW 10 Solutions Cosine Wave The graph shows the position x of an oscillating object as a function of time t. The equation of the graph is ( )()cosxt A t= ω +φ , where A is the amplitude, ω is the angular frequency, and φ is a phase at t = 0. The quantities M, N, and T are measurements to be used in your answers. a) What is A in the equation? M b) What is ω in the equation? 2 2 f T π ωπ== c) What is φ in the equation? () ()0cos 2 N xA T φ φπ== Energy of a Spring An object of mass m attached to a spring of force constant k oscillates with simple harmonic motion. The maximum displacement from equilibrium is A and the total mechanical energy of the system is E. a) What is the system's potential energy when its kinetic energy is equal to (3/4)E? () () () () () () () () () () 2 22 22 2 2 2 2 22 2 2 2 111 1 222 11 sin cos / 1 sin cos 2 E Kt Ut mv t kx t mxt kx t mA t kA t km kA t kA t kA ωωφ ωφω ωφ ωφ ⎡⎤=+= + = + ⎣⎦ ⎡⎤=+++= ⎣⎦ =+++= () () () 22 000 33 11 48 48 Kt E kA Pt E Kt E kA== =− == This happens at the phase: () () () () () 22 2 2 2 0 0 002 0 131 sin cos 24 sin 3tan 3 6 cos kAtEkAtE t tt t ωφ ωφ ωφ ωφ ωφ ωφ += += + =+=+ + 1 4 = 1 b) What is the object's velocity when its potential energy is (2/3)E? () () 22222 00 21 11 1 1 1 33 36 6 2 3 3 kk Pt E kA Kt E kA kA mv v A mm 2 Aω ω ⎡⎤ == == = =± ===± ⎢⎥ ⎣⎦ Mass and Simple Harmonic Motion Conceptual Question The shaker cart, shown in the figure, is the latest extreme sport craze. You stand inside of a small cart attached to a heavy-duty spring, the spring is compressed and released, and you shake back and forth, attempting to maintain your balance. Note that there is also a sandbag in the cart with you. At the instant you pass through the equilibrium position of the spring, you drop the sandbag out of the cart onto the ground. a) What effect does dropping the sandbag out of the cart at the equilibrium position have on the amplitude of your oscillation? By dropping the bag, you reduced the mass of the “oscillator” and its kinetic energy: at equilibrium the sandbag has kinetic energy, which leaves the system with the sandbag. On the other hand, at the equilibrium position, K=E (P=0), so you’ve reduced the total energy of the “oscillator”. This kinetic energy is transformed into the potential energy at the maximum displacement from the equilibrium position (x = ±A). Since the amplitude is related to the energy of the system by the equation 2 1 2 E kA= , a reduction in the energy of the system must lead to a reduction in the amplitude of the motion. b) What effect does dropping the sandbag out of the cart at the equilibrium position have on the maximum speed of the cart? Dropping the sandbag doesn't change the speed of the cart. Since energy doesn't leave the system at any time after the bag is dropped, the cart will always return to the equilibrium point with the same kinetic energy, and therefore the same maximum speed. Instead of dropping the sandbag as you pass through equilibrium, you decide to drop the sandbag when the cart is at its maximum distance from equilibrium. 2 c) What effect does dropping the sandbag at the cart’s maximum distance from equilibrium have on the amplitude of your oscillation? At the point of maximum displacement, all of the system's energy is stored in the spring. Therefore, dropping the sandbag has no effect on the system's total energy E. Since 2 1 2 E kA= , the amplitude cannot change. d) What effect does dropping the sandbag at the cart’s maximum distance from equilibrium have on the maximum speed of the cart? The system has not changed energy, but its mass has decreased. Since 2 max 1 2 E mv= , the maximum speed must increase to balance out the decreased mass. Simple Harmonic Motion Conceptual Question An object of mass m is attached to a vertically oriented spring. The object is pulled a short distance below its equilibrium position and released from rest. Set the origin of the coordinate system at the equilibrium position of the object and choose upward as the positive direction. Assume air resistance is so small that it can be ignored. Refer to these graphs when answering the following questions. a) Beginning the instant the object is released, select the graph that best matches the position vs. time graph for the object. H b) Beginning the instant the object is released, select the graph that best matches the velocity vs. time graph for the object. E c) Beginning the instant the object is released, select the graph that best matches the acceleration vs. time graph for the object. F 3 Vertical Mass-and-Spring Oscillator A block of mass m is attached to the end of an ideal spring. Due to the weight of the block, the block remains at rest when the spring is stretched a distance h from its equilibrium length. The spring has an unknown spring constant k. a) What is the spring constant k? Express the spring constant in terms of given quantities and g, the magnitude of the acceleration due to gravity. In equilibrium, mg kh mg k h == b) Suppose that the block gets bumped and undergoes a small vertical displacement. Find the resulting angular frequency ω of the block's oscillation about its equilibrium position. Express the frequency in terms of given quantities and g, the magnitude of the acceleration due to gravity. kg mh ω== It may seem that this result for the frequency does not depend on either the mass of the block or the spring constant, which might make little sense. However, these parameters are what would determine the extension h of the spring when the block is hanging: mg h k = . One way of thinking about this problem is to consider both k and g as unknowns. By measuring h and ω (both fairly simple measurements), and knowing the mass, you can determine the value of the spring constant and the acceleration due to gravity experimentally. Exercise 13.1 A piano string sounds a middle A by vibrating primarily at 220 Hz. a) Calculate the string's period. IDENTIFY and SET UP: The target variables are the period T and angular frequency .ω We are given the frequency f, so we can find these using Eqs.(13.1) and (13.2) EXECUTE: (a) 220 Hzf = 3 1/ 1/220 Hz 4.54 10 sTf − == = × 2 2 (220 Hz) 1380 rad/sfω ππ== = b) Calculate the string's angular frequency. 2(220 Hz) 440 Hzf == c) Calculate the period for a soprano singing an A one octave higher, which is twice the frequency of the piano string. 4 3 1/ 1/440 Hz 2.27 10 sTf − == = × (smaller by a factor of 2) d) Calculate the angular frequency for a soprano singing an A one octave higher, which is twice the frequency of the piano string. 2 2 (440 Hz) 2760 rad/sfω ππ== = (factor of 2 larger) EVALUATE: The angular frequency is directly proportional to the frequency and the period is inversely proportional to the frequency. Exercise 13.4 The displacement of an oscillating object as a function of time is shown in the figure. a) What is the frequency? IDENTIFY: The period is the time for one cycle and the amplitude is the maximum displacement from equilibrium. Both these values can be read from the graph. SET UP: The maximum x is 10.0 cm. The time for one cycle is 16.0 s. EXECUTE: (a) T 16.0 s= so 1 f T ==0.0625 Hz . b) What is the amplitude? 10.0 cmA = . c) What is the period? (c) 16.0 sT = d) What is the angular frequency of this motion? (d) 2 0.393 rad/sfω π== EVALUATE: After one cycle the motion repeats. Exercise 13.18 A 0.500-kg mass on a spring has velocity as a function of time given by () ( ) () 1 3.60 / sin 4.71 / 2 x vt cms s tπ − ⎡⎤ =− ⎣⎦ . a) What is the period? IDENTIFY: The general expression for vtis ( ) x ( ) sin( ) x vt A tω ω φ=−+. We can determine ω and A by comparing the equation in the problem to the general form. SET UP: 4.71 rad/sω = . 3.60 cm/s 0.0360 m/sAω == . EXECUTE: (a) 22 rad 1.33 s 4.71 rad/s T ππ ω == = b) What is the amplitude? 5 (b) 3 0.0360 m/s 0.0360 m/s 7.64 10 m 7.64 mm 4.71 rad/s A ω − ===×= c) What is the maximum acceleration of the mass? (c) 223 max (4.71 rad/s) (7.64 10 m) 0.169 m/saAω − == × = 2 d) What is the force constant of the spring? (d) k m ω= so . 22 (0.500 kg)(4.71 rad/s) 11.1 N/mkmω== = EVALUATE: The overall positive sign in the expression for and the factor of ( ) x vt / 2π− both are related to the phase factor φ in the general expression. Exercise 13.43 A building in San Francisco has light fixtures consisting of small 2.35-kg bulbs with shades hanging from the ceiling at the end of light thin cords 1.50 m long. If a minor earthquake occurs, how many swings per second will these fixtures make? IDENTIFY: Since the cord is much longer than the height of the object, the system can be modeled as a simple pendulum. We will assume the amplitude of swing is small, so that 2 L T g π= . SET UP: The number of swings per second is the frequency 11 2 g f TLπ == . EXECUTE: 2 19.80 m/s 0.407 swings per second 21.50 m f π == . EVALUATE: The period and frequency are both independent of the mass of the object. 6 Mike Gershenson HW 1 Solutions

Advertisement

#### Words From Our Students

"StudyBlue is great for studying. I love the study guides, flashcards, and quizzes. So extremely helpful for all of my classes!"

Alice, Arizona State University