Physics
Kinetic Energy in Simple Harmonic Motion
In simple harmonic motion, kinetic energy is the energy possessed by an object due to its motion. As the object oscillates back and forth, its kinetic energy varies, reaching a maximum at the equilibrium position and decreasing to zero at the extremes of the motion. This energy transformation is a key aspect of understanding the behavior of systems in simple harmonic motion.
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10 Key excerpts on "Kinetic 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(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, 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.
- David Halliday, Robert Resnick, Jearl Walker(Authors)
- 2023(Publication Date)
- Wiley(Publisher)
As position changes, the energy shifts between the two types, but the total is constant. FIGURE 15.2.1 (a) Potential energy U(t), kinetic energy K(t), and mechanical energy E as functions of time t for a linear harmonic oscillator. Note that all energies are positive and that the potential energy and the kinetic energy peak twice during every period. (b) Potential energy U(x), kinetic energy K(x), and mechanical energy E as functions of position x for a linear harmonic oscillator with amplitude x m . For x = 0 the energy is all kinetic, and for x = ±x m it is all potential. CHECKPOINT 15.2.1 In Fig. 15.1.7, the block has a kinetic energy of 3 J and the spring has an elastic poten- tial energy of 2 J when the block is at x = +2.0 cm. (a) What is the kinetic energy when the block is at x = 0? What is the elastic potential energy when the block is at (b) x = −2.0 cm and (c) x = −x m ? of time and then as a function of the oscillator’s position. 15.2.4 For a spring–block oscillator, determine the block’s position when the total energy is entirely kinetic energy and when it is entirely potential energy. 15.2 Energy in Simple Harmonic Motion 421 SAMPLE PROBLEM 15.2.1 SAMPLE PROBLEM 15.2.2 SHM potential energy, kinetic energy, mass dampers Energies of a linear oscillator Many tall buildings have mass dampers, which are anti- sway devices to prevent them from oscillating in a wind. The device might be a block oscillating at the end of a spring and on a lubricated track. If the building sways, say, eastward, the block also moves eastward but delayed enough so that when it finally moves, the building is then moving back westward. Thus, the motion of the oscilla- tor is out of step with the motion of the building. Suppose the block has mass m = 2.72 × 10 5 kg and is designed to oscillate at frequency f = 10.0 Hz and with amplitude x m = 20.0 cm.
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 - PDFFundamentals of Physics
Mechanics, Relativity, and Thermodynamics
- R. Shankar(Author)
- 2014(Publication Date)
- Yale University Press(Publisher)
The acceleration also oscillates but with an amplitude ω 2 A . These two results are true for any phase φ . 280 Simple Harmonic Motion Let us explicitly verify the law of conservation of energy. Consider the total energy: E ( t ) = 1 2 mv 2 + 1 2 kx 2 (17.15) = 1 2 m ω 2 A 2 sin 2 ω t + 1 2 kA 2 cos 2 ω t (17.16) = 1 2 kA 2 because ω 2 = k m . (17.17) Thus, by magic, the time-dependent terms sin 2 ω t and cos 2 ω t have the same coefficient, and you find that E ( t ) actually does not depend on time at all. Even though position and velocity are constantly changing, this combination will not depend on time. At the instant when the mass has reached one extremity and is about to swing back, it has no veloc-ity; it only has an x = A , and the energy of the oscillator is all potential energy, 1 2 kA 2 . 17.1 More examples of oscillations If a body is in stable equilibrium, and you disturb it, it rocks back and forth, executing simple harmonic motion. The standard textbook example is the mass on a spring, which we just studied. But it is a very generic situation, as shown in Figure 17.3. Skipping the mass-and-spring example, let us go the top right, where we have a beam hanging from the ceiling by a cable that is fixed to its center of mass (CM). If you twist it by an angle θ , it will try to untwist itself. Now we don’t have a restoring force but we have a restoring torque. What can be the expression for the restoring torque τ ? When you don’t do anything, the cable doesn’t do anything, so τ vanishes when θ = 0. If θ = 0, it is some function of θ , and the leading term in the Taylor expansion would be proportional to θ : τ ( θ ) = − κθ . (17.18) The coefficient κ is the torsion constant , and the minus sign tells you it’s a restoring torque. That means if you make θ positive, the torque will try to twist you the other way.
eBook - PDF- Stephen Lee(Author)
- 2014(Publication Date)
- CRC Press(Publisher)
a a O displacement time the amplitude, a the period, T 2 T T 306 AN INTRODUCTION TO MATHEMATICS FOR ENGINEERS : MECHANICS A physical model of an oscillating system A particle, P, of mass m kg is attached to a light, perfectly elastic spring which has stiffness k Nm 1 . The other end, E, of the spring is attached to a fixed point and P moves horizontally. The spring is its natural length when P is at the fixed point O. The particle is pulled aside to a point A where OA a m and let go. Figure 14.2 Now think about energy. At A, the spring has elastic energy 1 2 ka 2 and the particle has zero kinetic energy. The particle first comes to rest again at B so the elastic energy at B must also be 1 2 ka 2 and the length OB am. This is the amplitude of the motion. When the extension of the spring is x m and P has speed v ms 1 , the total energy is 1 2 mv 2 1 2 kx 2 so the principle of conservation of energy gives 1 2 mv 2 1 2 kx 2 1 2 ka 2 ⇒ v 2 m k ( a 2 x 2 ). This gives the speed of the particle in any position, even when x is negative, providing Hooke’s law still holds. E Spring stiffness , k Nm 1 Particle mass, m kg The particle is stationary at A B O a m A P P P P a m P The particle is stationary again at B SIMPLE HARMONIC MOTION 307 Q UESTION 14.1 How can this be done so that the only horizontal force is the force in the spring? Q UESTION 14.2 What happens to P as it moves from A to B? Then what happens? Q UESTION 14.3 If you could fix a pen at P and let it draw on a strip of paper moving at a steady rate perpendicular to EA (up the page), what would be the result? A mathematical model for simple harmonic motion At the instant the particle, P, is x m to the right of O the forces acting on it are as shown in figure 14.3. Figure 14.3 Its horizontal equation of motion is T mx .. By Hooke’s law, T kx so kx mx .. x .. m k x or d d 2 t x 2 m k x In equation , k and m are both positive constants so m k can be written as one positive constant.
eBook - PDF- H. W. Harkness(Author)
- 2014(Publication Date)
- Academic Press(Publisher)
When the vibrating body is at its extreme dis-placement it has no motion so the energy is all potential and should be equal to the total energy. At the instant it passes through the position of equilibrium the motion is a maximum and there is no resultant force acting, so the energy is all kinetic. At any intermediate point the system has both kinetic and poten-tial energy. The velocity at any displacement, x ft (Figure IV-7) is given by (IV-2) so the kinetic energy at this displacement is 1 E w *{V(A*-x*) } 2 ft-lb 2 g The potential energy at the same displacement is found by calculating the work to displace the body x ft. The force is variable being equal to (W/g)oe 2 x from (IV-15). As in all cases where work done against a varying force is calculated one considers the element of work done at any displacement r ft when the displacement increases by an amount dr ft so small that the 112 IV. UNDAMPED SIMPLE HARMONIC MOTION ■Φ-TT Equilibrium FIG. IV-7 force during this displacement is constant. The element of work for this increment of displacement is, dE = F r dr W g W dE = —oe 2 rdr g These elements of work are to be added up from the displace-ment zero to displacement x that is, W E = — co 2 I rdr W 9 r x g Jo 1 W E = ± — ω 2 * 2 ft-lb 2 g (IV-16) Notice that the potential energy of the system is proportional to the square of the displacement. ALTERNATIVE CALCULATION 113 It is to be noted that: (i) The expression contains only constants of the system and a numerical coefficient. (ii) The expression gives the maximum kinetic energy—the maximum velocity is ωΑ fps. (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.
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
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