Physics
Energy in Simple Harmonic Motion
In simple harmonic motion, energy is constantly exchanged between kinetic energy (energy of motion) and potential energy (stored energy). At the maximum displacement from the equilibrium position, all the energy is potential, while at the equilibrium position, all the energy is kinetic. As the object oscillates, the energy continuously oscillates between these two forms.
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10 Key excerpts on "Energy in Simple Harmonic Motion"
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 - 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, 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- H. W. Harkness(Author)
- 2014(Publication Date)
- Academic Press(Publisher)
(iii) The expression gives the maximum potential energy— expression (IV-16) when x = A. (iv) The energy of a system vibrating with SHM is propor-tional to the square of the amplitude. An Alternative Calculation of the Potential Energy If we knew the average force, F, during the displacement the work done and therefore the potential energy at the displacement x would be given at once by Fx. When, as in this case, the force is a linear function of the displacement, the average force is the arithmetic mean of the forces at the beginning and at the end of the displacement process, that is, ^ o + ^ y » i y A l b 2 2 g The potential energy is therefore, i w , iw u 2 g 2 g as before. (IV-17) or We may now write the expression for the total energy, 114 IV. UNDAMPED SIMPLE HARMONIC MOTION ILLUSTRATIVE EXAMPLES 1. A weight of 4 lb oscillates with SHM with an amplitude of 6 in. and a period of 0.005 sec. Calculate the total energy of the vibrating system. W 1 4 4π 2 ΤΕ = 2 7 ω 2 ^ 2 Χ 3 2 Χ Μ Χ ( ° · 5 > 2 = 2.46 ft-lb 2. The amplitude of vibration of an 8-lb weight which is hanging on the end of a certain spring is 4 in., and at this ampli-tude the rate of change of energy with amplitude is found to be 12 ft-lb/ft. Find the period of the motion and the force constant of the spring. 1 W 2 g :.^τ = —ω*Α = 12 ft-lb/ft dA g . _2_xj = 12 x 32 x 12 WA 8 x 4 144 lSeC > therefore, and T = ^ = ^ = 0.523 sec ω 12 π w w 2 g g F W 8 k = - — = — ω* - ^ x 144 = 36 lb/ft * g 32 The Form of the Expression for the Period of SHM In the case of the mass on the end of a spring we found that the period was given by, 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.
eBook - PDF- David Halliday, Robert Resnick, Jearl Walker(Authors)
- 2020(Publication Date)
- Wiley(Publisher)
For any angle α, cos 2 α + sin 2 α = 1. Thus, the quantity in the square brackets above is unity and we have E = U + K = 1 2 kx 2 m . (15-21) The mechanical energy of a linear oscillator is indeed constant and independent of time. The potential energy and kinetic energy of a linear oscillator are shown as functions of time t in Fig. 15-8a and as functions of displacement x in Fig. 15-8b. In any oscillating system, an element of springiness is needed to store the potential energy and an element of inertia is needed to store the kinetic energy. An Angular Simple Harmonic Oscillator Figure 15-9 shows an angular version of a simple harmonic oscillator; the element of springiness or elasticity is associated with the twisting of a suspension wire rather than the extension and compression of a spring as we previously had. The device is called a torsion pendulum, with torsion referring to the twisting. If we rotate the disk in Fig. 15-9 by some angular displacement θ from its rest position (where the reference line is at θ = 0) and release it, it will oscillate about that position in angular simple harmonic motion. Rotating the disk through an angle θ in either direction introduces a restoring torque given by τ = −κ θ. (15-22) Here κ (Greek kappa) is a constant, called the torsion constant, that depends on the length, diameter, and material of the suspension wire. Comparison of Eq. 15-22 with Eq. 15-10 leads us to suspect that Eq. 15-22 is the angular form of Hooke’s law, and that we can transform Eq. 15-13, which gives the period of linear SHM, into an equation for the period of angular SHM: We replace the spring constant k in Eq. 15-13 with its equivalent, the constant Additional examples, video, and practice available at WileyPLUS 15-3 AN ANGULAR SIMPLE HARMONIC OSCILLATOR Learning Objectives After reading this module, you should be able to . . . 15.23 Describe the motion of an angular simple harmonic oscillator.
eBook - ePub- H. John Pain(Author)
- 2013(Publication Date)
- Wiley(Publisher)
Figure 1.4 shows the distribution of energy versus displacement for simple harmonic motion. Note that the potential energy curveis parabolic with respect to x and is symmetric about x = 0, so that energy is stored in the oscillator both when x is positive and when it is negative, e.g. a spring stores energy whether compressed or extended, as does a gas in compression or rarefaction. The kinetic energy curveis parabolic with respect to both x and . The inversion of one curve with respect to the other displays the π /2 phase difference between the displacement (related to the potential energy) and the velocity (related to the kinetic energy).For any value of the displacement x the sum of the ordinates of both curves equals the total constant energy E .Figure 1.4Parabolic representation of potential energy and kinetic energy of simple harmonic motion versus displacement. Inversion of one curve with respect to the other shows a 90° phase difference. At any displacement value the sum of the ordinates of the curves equals the total constant energy E(Problems 1.10, 1.11, 1.12)Simple Harmonic Oscillations in an Electrical System
So far we have discussed the simple harmonic motion of the mechanical and fluid systems of Figure 1.1 , chiefly in terms of the inertial mass stretching the weightless spring of stiffness s . The stiffness s of a spring defines the difficulty of stretching; the reciprocal of the stiffness, the compliance C (where s = 1/C ) defines the ease with which the spring is stretched and potential energy stored. This notation of compliance C is useful when discussing the simple harmonic oscillations of the electrical circuit of Figure 1.1(h) and Figure 1.5 , where an inductance L is connected across the plates of a capacitance C . The force equation of the mechanical and fluid examples now becomes the voltage equation (balance of voltages) of the electrical circuit, but the form and solution of the equations and the oscillatory behaviour of the systems are identical.Figure 1.5Electrical system which oscillates simple harmonically. The sum of the voltages around the circuit is given by Kirchhoff’s law as L dI/dt + q /C
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
- David Halliday, Robert Resnick, Jearl Walker(Authors)
- 2023(Publication Date)
- Wiley(Publisher)
When an arrow is shot from a bow, the feathers at the end of the arrow manage to snake around the bow staff without hitting it because the arrow oscillates. When a coin drops into a metal collection plate, the coin oscillates with such a familiar ring that the coin’s denomination can be determined from the sound. When a rodeo cowboy rides a bull, the cow- boy oscillates wildly as the bull jumps and turns (at least the cowboy hopes to be oscillating). The study and control of oscillations are two of the primary goals of both physics and engineering. In this chapter we discuss a basic type of oscillation called simple harmonic motion. Heads Up. This material is quite challenging to most students. One reason is that there is a truckload of definitions and symbols to sort out, but the main reason is that we need to relate an object’s oscillations (something that we can see or even experience) to the equations and graphs for the oscillations. Relating the real, vis- ible motion to the abstraction of an equation or graph requires a lot of hard work. Simple Harmonic Motion Figure 15.1.1 shows a particle that is oscillating about the origin of an x axis, repeatedly going left and right by iden- tical amounts. The frequency f of the oscillation is the number of times per sec- ond that it completes a full oscillation (a cycle) and has the unit of hertz (abbrevi- ated Hz), where 1 hertz = 1 Hz = 1 oscillation per second = 1 s −1 . (15.1.1) The time for one full cycle is the period T of the oscillation, which is T = 1 __ f . (15.1.2) Any motion that repeats at regular intervals is called periodic motion or har- monic motion. However, here we are interested in a particular type of periodic motion called simple harmonic motion (SHM). Such motion is a sinusoidal func- tion of time t. That is, it can be written as a sine or a cosine of time t. Here we arbitrarily choose the cosine function and write the displacement (or position) of the particle in Fig.
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- 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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