Physics
Mechanical Energy in Simple Harmonic Motion
In simple harmonic motion, mechanical energy is the sum of kinetic and potential energy and remains constant throughout the motion. As the object oscillates, the energy is continuously converted between kinetic and potential forms. At the maximum displacement, the kinetic energy is zero and the potential energy is at its maximum, and vice versa at the equilibrium position.
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11 Key excerpts on "Mechanical Energy in Simple Harmonic Motion"
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler, Heath Jones, Matthew Collins, John Daicopoulos, Boris Blankleider(Authors)
- 2020(Publication Date)
- Wiley(Publisher)
The total mechanical energy E is the sum of these two energies: E = 0.414 J + 0.138 J = 0.552 J. Because the total mechanical energy remains constant during the motion, this value equals the initial total mechanical energy when the object is stationary and the energy is entirely elastic potential energy (E 0 = 1 2 kx 0 2 = 0.552 J). (b) When x 0 = 0.0450 m and x f = 0 m, we have v f = √ k m (x 2 0 - x 2 f ) = √ 545 N∕m 0.200 kg [(0.0450 m) 2 - (0 m) 2 ] = 2.35 m∕s Now the total mechanical energy is due entirely to the translational kinetic energy ( 1 2 mv 2 f = 0.552 J), since the elastic potential energy is zero (see part C of figure 10.17). Note that the total mechanical energy is the same as it is in solution part (a). In the absence of friction, the simple harmonic motion of a spring converts the different types of energy between one form and another, the total always remaining the same. FIGURE 10.17 The total mechanical energy of this system is entirely elastic potential energy (A), partly elastic potential energy and partly kinetic energy (B), and entirely kinetic energy (C). A B C x f = 0 cm x f x 0 0 = 0 m/s Conceptual example 8 takes advantage of energy conservation to illustrate what happens to the maximum speed, amplitude, and angular frequency of a simple harmonic oscillator when its mass is changed suddenly at a certain point in the motion. CONCEPTUAL EXAMPLE 8 Changing the mass of a simple harmonic oscillator Figure 10.18a shows a box of mass m attached to a spring that has a force constant k. The box rests on a horizontal, frictionless surface. The spring is initially stretched to x = A and then released from rest. The box executes simple harmonic motion that is characterised by a maximum speed v max x , an amplitude A, and an angular frequency . When the box is passing through the point where the spring is unstrained (x = 0 m), a second box of the same mass m and speed v max x is attached to it, as in part b of the drawing.
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2015(Publication Date)
- Wiley(Publisher)
The total mechanical energy E is the sum of these two energies: E 5 0.414 J 1 0.138 J 5 0.552 J. Because the total mechanical energy remains constant during the motion, this value equals the initial total mechanical energy when the object is stationary and the energy is entirely elastic potential energy (E 0 5 1 2 kx 2 0 5 0.552 J). (b) When x 0 5 0.0450 m and x f 5 0 m, we have v f 5 B k m (x 0 2 2 x f 2 ) 5 B 545 N/m 0.200 kg [(0.0450 m) 2 2 (0 m) 2 ] 5 2.35 m/s Now the total mechanical energy is due entirely to the translational kinetic energy ( 1 2 mv 2 f 5 0.552 J), since the elastic potential energy is zero (see part C of Figure 10.17). Note that the total mechanical energy is the same as it is in Solution part (a). In the absence of fric- tion, the simple harmonic motion of a spring converts the different types of energy between one form and another, the total always remaining the same. m m (a) (b) m x = 0 m x max x = A 0 = 0 m/s x max x max x = 0 m Figure 10.18 (a) A box of mass m, starting from rest at x 5 A, undergoes simple har- monic motion about x 5 0 m. (b) When x 5 0 m, a second box, with the same mass and speed, is attached. Conceptual Example 8 takes advantage of energy conservation to illustrate what hap- pens to the maximum speed, amplitude, and angular frequency of a simple harmonic oscil- lator when its mass is changed suddenly at a certain point in the motion. CONCEPTUAL EXAMPLE 8 | Changing the Mass of a Simple Harmonic Oscillator Figure 10.18a shows a box of mass m attached to a spring that has a force constant k. The box rests on a horizontal, frictionless surface. The spring is initially stretched to x 5 A and then released from rest. The box executes simple harmonic motion that is characterized by a 262 Chapter 10 | Simple Harmonic Motion and Elasticity maximum speed v x max , an amplitude A, and an angular frequency v.
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2021(Publication Date)
- Wiley(Publisher)
The total mechanical energy E is the sum of these two energies: E= 0.414 J + 0.138 J = 0.552 J. Because the total mechanical energy remains constant during the motion, this value equals the initial total mechanical energy when the object is stationary and the energy is entirely elastic potential energy ( E 0 = 1 _ 2 k x 0 2 = 0.552 J). (b) When x 0 = 0.0450 m and x f = 0 m, we have υ f = √ ___________ k __ m (x 0 2 − x f 2 ) = √ ___________________________ 545 N/m ________ 0.200 kg [(0.0450 m) 2 − (0 m) 2 ] = 2.35 m/s Now the total mechanical energy is due entirely to the transla- tional kinetic energy ( 1 _ 2 mυ f 2 = 0.552 J), since the elastic potential energy is zero (see part C of Interactive Figure 10.17). Note that the total mechanical energy is the same as it is in Solution part (a). In the absence of friction, the simple harmonic motion of a spring converts the different types of energy between one form and another, the total always remaining the same. INTERACTIVE FIGURE 10.17 The total mechanical energy of this system is entirely elastic potential energy (A), partly elastic potential energy and partly kinetic energy (B), and entirely kinetic energy (C). A B C x f = 0 cm x f x 0 v 0 = 0 m/s CONCEPTUAL EX AMPLE 8 Changing the Mass of a Simple Harmonic Oscillator Figure 10.18a shows a box of mass m attached to a spring that has a force constant k. The box rests on a horizontal, friction- less surface. The spring is initially stretched to x = A and then released from rest. The box executes simple harmonic motion that is characterized by a maximum speed υ x max , an amplitude A, and an angular frequency . When the box is passing through the point where the spring is unstrained (x = 0 m), a second box of the same mass m and speed υ x max , is attached to it, as in part b of the drawing. Discuss what happens to (a) the maximum speed, (b) the amplitude, and (c) the angular frequency of the subse- quent simple harmonic motion.
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2015(Publication Date)
- Wiley(Publisher)
232 Chapter 10 | Simple Harmonic Motion and Elasticity A B C x f = 0 cm x f x 0 v 0 = 0 m/s Figure 10.17 The total mechanical energy of this system is entirely elastic potential energy (A), partly elastic potential energy and partly kinetic energy (B), and entirely kinetic energy (C). m m (a) (b) m x = 0 m x max x = A 0 = 0 m/s x max x max x = 0 m Figure 10.18 (a) A box of mass m, starting from rest at x 5 A, undergoes simple har- monic motion about x 5 0 m. (b) When x 5 0 m, a second box, with the same mass and speed, is attached. As Section 6.5 discusses, the total mechanical energy is conserved when external non- conservative forces (such as friction) do no net work, that is, when W nc 5 0 J. Then, the final and initial values of E are the same: E f 5 E 0 . The principle of conservation of total mechanical energy is the subject of the next example. EXAMPLE 7 | An Object on a Horizontal Spring Figure 10.17 shows an object of mass m 5 0.200 kg that is vibrating on a horizontal friction- less table. The spring has a spring constant of k 5 545 N/m. The spring is stretched initially to x 0 5 4.50 cm and is then released from rest (see part A of the drawing). Determine the final translational speed v f of the object when the final displacement of the spring is (a) x f 5 2.25 cm and (b) x f 5 0 cm. Reasoning The conservation of mechanical energy indicates that, in the absence of friction (a nonconservative force), the final and initial total mechanical energies are the same: E f 5 E 0 1 2 mv f 2 1 1 2 I v f 2 1 mgh f 1 1 2 kx f 2 5 1 2 mv 0 2 1 1 2 Iv 0 2 1 mgh 0 1 1 2 kx 0 2 Since the object is moving on a horizontal table, the final and initial heights are the same: h f 5 h 0 . The object is not rotating, so its angular speed is zero: v f 5 v 0 5 0 rad/s. Also, as the problem states, the initial translational speed of the object is zero, v 0 5 0 m/s.
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2018(Publication Date)
- Wiley(Publisher)
A B C x f = 0 cm x f x 0 v 0 = 0 m/s INTERACTIVE FIGURE 10.17 The total mechanical energy of this system is entirely elastic potential energy (A), partly elastic potential energy and partly kinetic energy (B), and entirely kinetic energy (C). Conceptual Example 8 takes advantage of energy conservation to illustrate what happens to the maximum speed, amplitude, and angular frequency of a simple harmonic oscillator when its mass is changed suddenly at a certain point in the motion. CONCEPTUAL EXAMPLE 8 Changing the Mass of a Simple Harmonic Oscillator Figure 10.18a shows a box of mass m attached to a spring that has a force constant k. The box rests on a horizontal, frictionless surface. The spring is initially stretched to x = A and then released from rest. The box executes simple harmonic motion that is characterized by a maximum speed x max , an amplitude A, and an angular frequency . When the box is passing through the point where the spring is unstrained (x = 0 m), a second box of the same mass m and speed x max is attached to it, as in part b of the drawing. Discuss what happens to (a) the maximum speed, (b) the amplitude, and (c) the angular frequency of the subsequent simple harmonic motion. Reasoning and Solution (a) The maximum speed of an object in simple harmonic motion occurs when the object is passing through the point where the spring is unstrained (x = 0 m), as in Figure 10.18b. Since the second box is attached at this point with the same speed, the maximum speed of the two-box system remains the same as that of the one-box system. 10.3 Energy and Simple Harmonic Motion 269 In the previous two examples, gravitational potential energy plays no role because the spring is horizontal. The next example illustrates that gravitational potential energy must be taken into account when a spring is oriented vertically. (b) At the same speed, the maximum kinetic energy of the two boxes is twice that of a single box, since the mass is twice as much.
eBook - PDFWorkshop Physics Activity Guide Module 2
Mechanics II
- Priscilla W. Laws, David P. Jackson, Brett J. Pearson(Authors)
- 2023(Publication Date)
- Wiley(Publisher)
Simple harmonic motion involves a displacement that changes sinusoidally in time. We will study the behavior of two systems that undergo SHM—an object hanging from a spring (a mass-spring system) and a simple pendulum that oscillates at small angles. Pendula and masses on springs are two common examples of periodic systems that oscillate with SHM (or at least approximately SHM). Simple harmonic motion is quite common in the physical world and helps us describe such diverse phenomena as the behavior of the tiniest fundamental particles, how clocks work (both analog and digital), and the periodic signals emitted by pulsars. We will devise ways to describe oscillating systems in general and then apply these descriptions to SHM. Questions we will address include: What is periodic motion and how can it be characterized? What factors do the motions of a mass on a spring and a simple pendulum depend on? What mathematical behavior is required for a motion to be considered simple harmonic? How do Newton’s laws allow us to predict the motion of a mass on a spring or a pendulum oscillating at small angles? UNIT 14: SIMPLE HARMONIC MOTION 461 OSCILLATING SYSTEMS 14.2 CHARACTERISTICS OF PERIODIC SYSTEMS A mass on a spring, a simple pendulum, and a point on a wheel rotating with uniform rotational velocity undergo periodic motions that are quite similar (see Fig. 14.1). To observe the experiments in this section, you will need the following: • 1 pendulum bob • 1 string • 1 spring • 1 mass pan • 1 mass set, 100 g, 200 g, 500 g, 1 kg, 2 kg, etc. • 1 rotating disk, with a pin (or some other marker) on its outer rim • 1 variable speed motor (to drive the disk) • 1 table clamp • 2 rods • 1 right angle clamp • 1 stopwatch (or use the one on your phone) • 1 ruler • 1 protractor XII VI III IV V VII VIII IX X XI II I Fig. 14.1. A pendulum bob, a mass on a spring, and a peg on a rotating disk as oscillating objects.
eBook - PDF- H. W. Harkness(Author)
- 2014(Publication Date)
- Academic Press(Publisher)
USEFUL EXPRESSION IN TERMS OF ENERGY 115 The numerator of the expression under the root sign is the mass of the vibrating body and the denominator is the force constant, that is, the restoring force per unit displacement. In the case of the simple pendulum, the mass of the bob is Wjg slugs and we saw that the restoring force was {Wjl) · x so that the restoring force per unit displacement is Wjl. Hence, the period of the simple pendulum is An electric circuit with capacitance and inductance in series is analogous to a mechanical system, such as a mass hanging on a spring, and will oscillate with SHM. In this case if the electrons in the circuit are displaced from their equilibrium configuration, (the capacitor is charged) they will subsequently oscillate with SHM in the circuit. If the inductance of the circuit is L henrys, the capacitance C farads then the period of these oscillations is given by, Inductance in the electric circuit is analogous to mass in the mechanical system and the reciprocal of capacitance is analogous to the force constant of the mechanical system. In general, the period of a system vibrating with SHM (un-damped) is given by 2π times the square root of the ratio of the inertia of the system and its force constant. A Useful Expression for the Period in Undamped Harmonic Motion in Terms of the Energy Let the kinetic energy for unit velocity of the system be A and the potential energy for unit displacement of the system be B. Then, since the total energy, TE> of the system is constant we write Av 2 + Bx 2 = constant 116 IV. UNDAMPED SIMPLE HARMONIC MOTION so 2Av dv + IBx dx = 0 or since dv dv v-j-= — dx dt then This defines simple harmonic motion, and B/A is ω 2 , so at once we write T= 2π^ (IV-17b) Period for a Mass Hanging on a Heavy Spring Thus far we have assumed that the mass of the spring sup-porting the oscillating mass itself was negligible.
- H. D. Ram, A. K. Chauhan(Authors)
- 2015(Publication Date)
- Cambridge University Press(Publisher)
ϕ 0 π/ 2 π 3 π/ 2 2 π x 0 A 0 -A 0 dotnosp x A ω 0 -A ω 0 A ω dotnospdotnosp x 0 -A ω 2 0 A ω 2 0 9.3 Differential equation governing the simple harmonic motion Differentiating the equation 9.1 twice with respect to t , we get d x dt A t 2 2 2 = -+ ( ) ω ω α sin d x dt x 2 2 2 = -ω Mechanical Vibration | 487 dotnospdotnosp x x + = ω 2 0 (9.2) Solution of the equation 9.2 is given by the equation 9.1 which is repeated as follows: x A t = + ( ) sin ω α (9.1) 9.4 Elements of vibrating systems Inertia (mass) and elasticity are the physical properties of matter. These two properties are the main requirements of vibrating system. The inertia is the property of matter, which tries to maintain the motion of the body, and the elasticity is the property of matter, which tries to oppose the change in shape or size. Inertia and elasticity are inseparable properties of matter, i.e., we cannot have mass without elasticity or elasticity without mass. But for the sake of physical modelling of the system these are separated. The elasticity of a system is modelled by a massless spring of stiffness, k and inertia of the system is modelled by a rigid body having mass, m only. A physical model of vibrating system is shown in Figure 9.1. Figure 9.1 9.5 Free vibration When a vibrating system is disturbed from its equilibrium position and left, the resulting motion is called free vibration. 9.6 Mathematical model representing the ideal system A particle is said to be in equilibrium if acceleration of the particle is zero. Acceleration of the particle will be zero only when the resultant force on it is zero. The force on mass is zero when the spring is undeformed, i.e., the displacement, x = 0 . This position is called static equilibrium position. Suppose at instant, t , the displacement of mass from static equilibrium position is x and velocity is dotnosp x . The total energy of the system is: E k x m x = + 1 2 1 2 2 2 dotnosp
eBook - PDF- John Matolyak, Ajawad Haija(Authors)
- 2013(Publication Date)
- CRC Press(Publisher)
The spring, compressed to its maximum, exerts a force on m to the right, pushing it back toward the equilibrium position, passing it and stretching the spring further to R as shown in (d). 179 Simple Harmonic Motion © 2010 Taylor & Francis Group, LLC toward the equilibrium position. Again, the mass gradually gaining momentum passes this position and stretches the spring to its maximum at R (Figure 9.4d), where one complete oscillation from R to L to R has been executed. This oscillation is also called a cycle, and as the system moves back and forth in sustained cycles, it is executing an SHM. The displacement from O to R through which m was pulled prior to releasing it is called the amplitude of the oscillation and is denoted by A. 9.3.1 K INETIC E NERGY AND T OTAL M ECHANICAL E NERGY OF M ASS -S PRING S YSTEM IN SHM It has been established in Chapter 6 that the total mechanical energy E of an object acted upon by conservative forces is conserved. As the spring force is conservative, then at any position, x, the mass-spring system has a total mechanical energy E 1 2 mv 1 2 kx 2 2 = + (9.5) that retains the same value. Hence, in Figure 9.4, equating the system’s total energy at point R to its value at the origin O, gives E R = E O , or 1 2 mv 1 2 kx 1 2 mv 1 2 kx 2 R 2 R 2 o 2 o . = + + Let OR = A. As the block’s velocity at R is zero, the above equation reduces to 0 1 2 kA 1 2 mv 0. 2 2 o + = + Also, as m has its maximum velocity at O, the above equation becomes 1 2 kA 1 2 mv 2 max 2 = (9.6) or v k m A A , max = ± = ±ω (9.7) where ω = ± k m (9.8) is the angular frequency of the block. The plus or minus signs in Equation 9.8 signify the directions of the movement of the block as it goes through the equilibrium position. The positive sign for v max indicates that the velocity of the block at the point x = 0 is moving toward the positive x axis chosen
eBook - PDF- Michael Tammaro(Author)
- 2019(Publication Date)
- Wiley(Publisher)
Now let’s compare the periods of the simple and physical pendulums in Figure 11.4.1. For the simple pendulum, we use Equation 11.4.2: T d g 2π = The moment of inertia of a uniform rod is listed in Table 8.5.1: I m d md 2 1 3 2 4 3 2 ( ) = = Substituting this into Equation 11.4.4, we have T md mgd d g 2 2 4 3 4 3 2 π π = = The physical pendulum’s period, therefore, is longer than the simple pendulum’s period by a factor of 4 3 1.15 ≈ / . Note that the period of the physical pendulum is also independent of its total mass. Figure 11.4.1 A simple pendulum (left) and a physical pendulum (right). Center of mass Simple pendulum Physical pendulum Pivot point (axis) d d mg mg 11.5 Solve problems dealing with energy in oscillating systems. The energy stored in a compressed or stretched spring is U kx 1 2 2 = (Equation 6.4.4), where k is the spring constant, x is the amount of stretch or compression, and x 0 = represents no compression or stretch. Thus, for a mass oscillating back and forth on a horizontal surface 11.5 ENERGY IN SIMPLE HARMONIC MOTION Learning Objective I N T E R A C T I V E F E A T U R E 298 | Chapter 11 from the end of a spring, the total mechanical energy is given by E mv kx tot 1 2 2 1 2 2 = + (11.5.1) If there is no friction, then mechanical energy is conserved and the total mechanical energy does not change. Animated Figure 11.5.1 illustrates this situation. Animated Figure 11.5.1 As the mass oscillates back and forth, the energy is transformed from kinetic to potential to kinetic, etc., but the total energy does not change. I N T E R A C T I V E F E A T U R E The speed has its maximum value at the equilibrium position, x 0 = , where the potential energy stored in the spring is zero and all of the energy is kinetic energy. The velocity is zero at x A = ± , where the kinetic energy is zero and all of the energy is potential energy stored in the spring.
eBook - PDF- David Agmon, Paul Gluck;;;(Authors)
- 2009(Publication Date)
- WSPC(Publisher)
Indicate on your graph the amplitude of motion A. Solution The continuous parabola represents the elastic potential energy U(x) = he 1 12, the dotted parabola the kinetic energy E k = E - kx 12, the horizontal dotted line is E= U + E k . The amplitude x = ±A is obtained by dropping vertically from 250 Classical and Relativistic Mechanics points P and Q to the x-axis, and . The graph illustrates the fact that for -A < x < A, the total energy U + E± = const. Simple dynamical arguments show that in any periodic motion (not just SHM) the maximum velocity, and with it the kinetic energy, is obtained when the body passes its equilibrium position. In the absence of dissipative forces this implies that the potential energy is a minimum there, since total energy is conserved. This ties in with our earlier notion that stable equilibrium is ensured by a restoring force. The latter is responsible for the periodic motion about the equilibrium point, which is 1 E = E ~~ i iV w) y 1 | i -A +A indeed one of minimum potential energy (see section 6.7 in chapter 6). u m rnmmm ^ ^ / ; , ^ A /-Example 7 Rope tied to a spring. A uniform rope of mass m and length I rests on a smooth horizontal table. One of its ends is tied to a spring offorce constant k fastened to a wall The system is in equilibrium when the spring is stretched by x 2 from its undistorted length and a length X of rope dangles from the table. From this state the rope is pulled from rest at its free end by a further length x and then released, {a) Write an expression for the resultant force on the rope as a function of x. (b) Derive conditions for the system to be in a state of stable, unstable or neutral equilibrium, (c) In which of these states will the system execute SHM, and what will be its period? Solution (a) In equilibrium the tension caused by the spring force kx 2 equals the weight of the dangling part of the rope mgx x ll , so that x 2 lx x = mglkl .
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