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“We are to admit no more causes of natural things than such as are both true and sufficient to explain their appearances.” Isaac Newton
13.1 Introduction to Periodic Motion
Periodic motion is any motion that repeats itself in equal intervals of time. The uniformly rotating earth represents a periodic motion that repeats itself every 24 hours. The motion of the earth around the sun is periodic, repeating itself every 12 months. A vibrating spring and a pendulum also exhibit periodic motion. The period of the motion is defined as the time for the motion to repeat itself. A special type of periodic motion is simple harmonic motion and we now proceed to investigate it.
13.2 Simple Harmonic Motion
An example of simple harmonic motion is the vibration of a mass m, attached to a spring of negligible mass, as the mass slides on a frictionless surface, as shown in figure 13.1. We say that the mass, in the unstretched position, figure 13.1(a), is in its equilibrium position. If an applied force FA acts on the mass, the mass will be displaced to the right of its equilibrium position a distance x, figure 13.1(b). The distance that the spring stretches, obtained from Hooke’s law, is
F A = kx
The displacement x is defined as the distance the body moves from its equilibrium position. Because F A is a force that pulls the mass to the right, it is also the force that pulls the spring to the right. By Newton’s third law there is an equal but opposite elastic force exerted by the spring on the mass pulling the mass toward the left. Since this force tends to restore the mass to its original position, we call it the restoring force F R. Because the restoring force is opposite to the applied force, it is given by F R = − F A = − kx (13.1)
When the applied force F A is removed, the elastic restoring force F R is then the only force acting on the mass m, figure 13.1(c), and it tries to restore m to its equilibrium position. We can then find the acceleration of the mass from Newton’s second law as
ma = F R = − kx Thus, a = − k x (13.2) m
Figure 13.1 The vibrating spring.
Equation 13.2 is the defining equation for simple harmonic motion. Simple harmonic motion is motion in which the acceleration of a body is directly proportional to its displacement from the equilibrium position but in the opposite direction. A vibrating system that executes simple harmonic motion is sometimes called a harmonic oscillator. Because the acceleration is directly proportional to the displacement x in simple harmonic motion, the acceleration of the system is not constant but varies with x. At large displacements, the acceleration is large, at small displacements the acceleration is small. Describing the vibratory motion of the mass m requires some new techniques and we will do so in section 13.3. However, let us first look at the motion from a physical point of view. The mass m in figure 13.2(a) is pulled a distance x = A to the right, and is then released. The maximum restoring force on m acts to the left at this position because
F Rmax = − kx max = − kA
The maximum displacement A is called the amplitude of the motion. At this position the mass experiences its maximum acceleration to the left. From equation
and hence a maximum acceleration
a max = − k (− A ) = k A m m
also to the right. The mass moves to the right while the force F R and the acceleration a decreases with x until x is again equal to zero, figure 13.2(d). Then F R and a are also zero. Because of the inertia of the mass, it moves past the equilibrium position to positive values of x. The restoring force again acts toward the left, slowing down the mass. When the displacement x is equal to A, figure 13.2(e), the mass momentarily comes to rest and then the motion repeats itself. One complete motion from x = A and back to x = A is called a cycle or an oscillation. The period T is the time for one complete oscillation, and the frequency f is the number of complete oscillations or cycles made in unit time. The period and the frequency are reciprocal to each other, that is,
f = 1 (13.3) T
The unit for a period is the second, while the unit for frequency, called a hertz, is one cycle per second. The hertz is abbreviated, Hz. Also note that a cycle is a number not a dimensional quantity and can be dropped from the computations whenever doing so is useful.
13.3 Analysis of Simple Harmonic Motion -- The
Reference Circle
As pointed out in section 13.2, the acceleration of the mass in the vibrating spring system is given by equation 13.2 as a = − k x m
Since the acceleration a = d^2 x / dt^2 , equation 13.2 can be written as
d^2 x = − k x (13.4) dt^2 m
Equation 13.4 is a second-order differential equation that completely describes the simple harmonic motion of the mass m. Unfortunately, the solution of such differential equations is beyond the scope of this course. We also can not use the kinematic equations derived in chapter 2 because they were based on the assumption that the acceleration of the system was a constant. As we can see from
equation 13.2, this assumption is no longer valid because the acceleration varies with the displacement x. Thus, a new set of equations must be derived to describe simple harmonic motion. It turns out that simple harmonic motion is related to the uniform circular motion studied in chapter 4. An analysis of uniform circular motion gives us a set of equations to describe simple harmonic motion. As an example, consider a point Q moving in uniform circular motion with an angular velocity ω, as shown in figure 13.3(a). At a particular instant of time t, the angle θ that Q has turned through is
θ = ω t (13.5)
Figure 13.3 Simple harmonic motion and the reference circle.
The projection of point Q onto the x -axis gives the point P. As Q rotates in the circle, P oscillates back and forth along the x -axis, figure 13.3(b). That is, when Q is at position 1, P is at 1. As Q moves to position 2 on the circle, P moves to the left along the x -axis to position 2 ’. As Q moves to position 3, P moves on the x -axis to position 3 ’ , which is of course the value of x = 0. As Q moves to position 4 on the circle, P moves along the negative x -axis to position 4 ’. When Q arrives at position 5, P is also there. As Q moves to position 6 on the circle, P moves to position 6 ’ on the x -axis. Then finally, as Q moves through positions 7, 8, and 1, P moves through 7 ’ , 8 ’ , and 1, respectively. The oscillatory motion of point P on the x -axis corresponds to
v = − V T sin θ (13.11)
The minus sign indicates that the velocity of P is toward the left at this position. The linear velocity V T of the point Q is related to the angular velocity ω by equation 9.3 of chapter 9, that is v = r ω
For the reference circle, v = V T and r is the amplitude A. Hence, the tangential velocity V T is given by V T = ω A (13.12)
Using equations 13.11, 13.12, and 13.5, the velocity of point P becomes
v = −ω A sin ω t (13.13)
Equation 13.13 is the second of the kinematic equations for simple harmonic motion and it gives the speed of the vibrating mass at any time t. We could also find the velocity equation 13.13 by a direct differentiation of equation 13.7 as v = dx = d ( A cos ω t ) = A (− sin ω t ) d (ω t ) dt dt dt and v = −ω A sin ω t
The differentiation is a simpler derivation of the velocity of the mass m but it does not show the physical relation between the reference circle and simple harmonic motion. A third kinematic equation for simple harmonic motion giving the speed of the vibrating body as a function of displacement can be found from equation 13. by using the trigonometric identity sin^2 θ + cos 2 θ = 1 or
sin ✕ = ± 1 − cos 2 ✕
From figure 13.3(a) or equation 13.6, we have
cos θ = x A Hence,
sin ✕ = ± 1 − x (13.14)
2 A^2
Substituting equation 13.14 back into equation 13.13, we get
v = ± ✬ A 1 − x
2 A^2 or v = ± ✬ A^2 − x^2 (13.15)
Equation 13.15 is the third of the kinematic equations for simple harmonic motion and it gives the velocity of the moving body at any displacement x. The ± sign in equation 13.15 indicates the direction of the vibrating body. If the body is moving to the right, then the positive sign (+) is used. If the body is moving to the left, then the negative sign (−) is used. Finally, we can find the acceleration of the vibrating body using the reference circle in figure 13.3(d). The point Q in uniform circular motion experiences a centripetal acceleration ac pointing toward the center of the circle in figure 13.3(d). The x -component of the centripetal acceleration is the acceleration of the vibrating body at the point P. That is, a = − a c cos θ (13.16)
The minus sign again indicates that the acceleration is toward the left. But recall from chapter 4 that the magnitude of the centripetal acceleration is
a c = v^2 (4.56) r
where v represents the tangential speed of the rotating object, which in the present case is V T , and r is the radius of the circle, which in the present case is the radius of the reference circle A. Thus, a c = V T^2 A
But we saw in equation 13.12 that V T = ω A, therefore
a c = ω^2 A
The acceleration of the mass m, equation 13.16, thus becomes
a = −ω^2 A cos ω t (13.17)
Equation 13.17 is the fourth of the kinematic equations for simple harmonic motion. It gives the acceleration of the vibrating body at any time t. Equation 13.17 could also have been found by a direct differentiation of equation 13.13 as
v = − ω A sin ω t (13.13) v = ± ✬ A^2 − x^2 (13.15) a = − ω^2 A cos ω t (13.17) F = − m ω^2 A cos ω t (13.18)
where, from equations 13.9 and 13.20, we have
✬ = 2 ✜ f = (^) mk
A plot of the displacement, velocity, and acceleration of the vibrating body as a function of time are shown in figure 13.4. We can see that the mathematical description follows the physical description in figure 13.2. When x = A, the
Figure 13.4 Displacement, velocity, and acceleration in simple harmonic motion.
maximum displacement, the velocity v is zero, while the acceleration is at its maximum value of −ω^2 A. The minus sign indicates that the acceleration is toward the left. The force is at its maximum value of − m ω^2 A, where the minus sign shows
that the restoring force is pulling the mass back toward its equilibrium position. At the equilibrium position x = 0, a = 0, and F = 0, but v has its maximum velocity of −ω A toward the left. As x goes to negative values, the force and the acceleration become positive, slowing down the motion to the left, and hence v starts to decrease. At x = − A the velocity is zero and the force and acceleration take on their maximum values toward the right, tending to restore the mass to its equilibrium position. As x becomes less negative, the velocity to the right increases, until it picks up its maximum value of ω A at x = 0, the equilibrium position, where F and a are both zero. Because of this large velocity, the mass passes the equilibrium position in its motion toward the right. However, as soon as x becomes positive, the force and the acceleration become negative thereby slowing down the mass until its velocity becomes zero at the maximum displacement A. One entire cycle has been completed, and the motion starts over again. (We should emphasize here that in this vibratory motion there are two places where the velocity is instantaneously zero, x = A and x = − A, even though the instantaneous acceleration is nonzero there.) Sometimes the vibratory motion is so rapid that the actual displacement, velocity, and acceleration at every instant of time are not as important as the gross motion, which can be described in terms of the frequency or period of the motion. We can find the frequency of the vibrating mass in terms of the spring constant k and the vibrating mass m by setting equation 13.9 equal to equation 13.20. Thus,
✬ = 2 ✜ f = (^) mk
Solving for the frequency f, we obtain
f = (^21) ✜ mk (13.21)
Equation 13.21 gives the frequency of the vibration. Because the period of the vibrating motion is the reciprocal of the frequency, we get for the period
T = 2 ✜ mk (13.22)
Equation 13.22 gives the period of the simple harmonic motion in terms of the mass m in motion and the spring constant k. Notice that for a particular value of m and k, the period of the motion remains a constant throughout the motion.
Example 13.
An example of simple harmonic motion. A mass of 0.300 kg is placed on a vertical spring and the spring stretches by 10.0 cm. It is then pulled down an additional 5. cm and then released. Find (a) the spring constant k, (b) the angular frequency ω, (c) the frequency f, (d) the period T, (e) the maximum velocity of the vibrating mass,
e. The maximum velocity, found from equation 13.13, is
v max = ω A = (9.90 rad/s)(5.00 × 10 −^2 m) = 0.495 m/s
f. The maximum acceleration, found from equation 13.17, is
a max = ω^2 A = (9.90 rad/s)^2 (5.00 × 10 −^2 m) = 4.90 m/s 2
g. The maximum restoring force, found from Hooke’s law, is
F max = kx max = kA = (29.4 N/m)(5.00 × 10 −^2 m) = 1.47 N
h. The velocity of the mass at x = 2.00 cm, found from equation 13.15, is
v = ± ✬ A^2 − x^2 v = ± (9.90 rad/s) (5.00 % 10 −^2 m)^2 − (2.00 % 10 −^2 m)^2 = ± 0.454 m/s
where v is positive when moving to the right and negative when moving to the left. i. The equation of the displacement at any instant of time, found from equation 13.7, is x = A cos ω t = (5.00 × 10 −^2 m) cos(9.90 rad/s) t
The equation of the velocity at any instant of time, found from equation 13.13, is
v = −ω A sin ω t = −(9.90 rad/s)(5.00 × 10 −^2 m)sin(9.90 rad/s) t = −(0.495 m/s)sin(9.90 rad/s) t
The equation of the acceleration at any time, found from equation 13.17, is
a = −ω^2 A cos ω t = −(9.90 rad/s)^2 (5.00 × 10 −^2 m)cos(9.90 rad/s) t = −(4.90 m/s 2 )cos(9.90 rad/s) t
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13.4 Conservation of Energy and the Vibrating Spring
The vibrating spring system of figure 13.2 can also be described in terms of the law of conservation of energy. When the spring is stretched to its maximum displacement A, work is done on the spring, and hence the spring contains potential energy. The mass m attached to the spring also has that potential energy. The kinetic energy is equal to zero at this point because v = 0 at the maximum displacement. The total energy of the system is thus
E tot = PE + KE = PE
But recall from chapter 7 that the potential energy of a spring was given by equation 7.25 as PE = 1 kx^2 2 Hence, the total energy is E tot = PE = 1 kA^2 (13.23) 2
When the spring is released, the mass moves to a smaller displacement x, and is moving at a speed v. At this arbitrary position x, the mass will have both potential energy and kinetic energy. The law of conservation of energy then yields
E tot = PE + KE E tot = 1 kx^2 + 1 mv^2 (13.24) 2 2
But the total energy imparted to the mass m is given by equation 13.23. Hence, the law of conservation of energy gives E tot = E tot 1 kA^2 = 1 kx^2 + 1 mv^2 (13.25) 2 2 2
We can also use equation 13.25 to find the velocity of the moving body at any displacement x. Thus, 1 mv^2 = 1 kA^2 − 1 kx^2 2 2 2 v^2 = k ( A^2 − x^2 ) m v =! (^) mk ( A^2 − x^2 ) (13.26)
When the oscillating mass reaches x = A, the kinetic energy becomes zero since
1 kA^2 = 1 kA^2 + 1 mv^2 2 2 2 1 mv^2 = 1 kA^2 − 1 kA^2 = 0 2 2 2 = KE = 0
As the oscillation continues there is a constant interchange of energy between potential energy and kinetic energy but the total energy of the system remains a constant.
Example 13.
Conservation of energy applied to a spring. A horizontal spring has a spring constant of 29.4 N/m. A mass of 300 g is attached to the spring and displaced 5.00 cm. The mass is then released. Find (a) the total energy of the system, (b) the maximum velocity of the system, and (c) the potential energy and kinetic energy for x = 2. cm.
Solution
a. The total energy of the system is E tot = 1 kA^2 2 = 1 (29.4 N/m)(5.00 × 10 −^2 m)^2 2 = 3.68 × 10 −^2 J
b. The maximum velocity occurs when x = 0 and the potential energy is zero. Therefore, using equation 13.28,
v (^) max = (^) m k A
v (^) max = (^) 3.0029.4 N/m % 10 − (^1) kg (5.00 % 10 −^2 m)
= 0.495 m/s
c. The potential energy at 2.00 cm is
PE = 1 kx^2 = 1 (29.4 N/m)(2.00 × 10 −^2 m)^2 2 2 = 5.88 × 10 −^3 J The kinetic energy at 2.00 cm is
KE = 1 mv^2 = 1 m k ( A^2 − x^2 ) 2 2 m = 1 (29.4 N/m)[(5.00 × 10 −^2 m)^2 − (2.00 × 10 −2 m) 2 ] 2 = 3.09 × 10 −^2 J
Note that the sum of the potential energy and the kinetic energy is equal to the same value for the total energy found in part a.
To go to this Interactive Example click on this sentence.
13.5 The Simple Pendulum
Another example of periodic motion is a pendulum. A simple pendulum is a bob that is attached to a string and allowed to oscillate, as shown in figure 13.6(a). The
Figure 13.6 The simple pendulum.
bob oscillates because there is a restoring force, given by
Restoring force = − mg sin θ (13.29)
This restoring force is just the component of the weight of the bob that is perpendicular to the string, as shown in figure 13.6(b). If Newton’s second law, F = ma, is applied to the motion of the pendulum bob, we get
− mg sin θ = ma
The tangential acceleration of the pendulum bob is thus
or k P = mg (13.34) l
Equation 13.34 states that the motion of a pendulum can be described by the equations developed for the vibrating spring by using the equivalent spring constant of the pendulum k p. Thus, the period of motion of the pendulum, found from equation 13.22, is
T p = 2 ✜ (^) km p
= 2 ✜ (^) mgm / l
T p = 2 ✜ (^) gl (13.35)
The period of motion of the pendulum is independent of the mass m of the bob but is directly proportional to the square root of the length of the string. If the angle θ is equal to 15^0 on either side of the central position, then the true period differs from that given by equation 13.35 by less than 0.5%. The pendulum can be used as a very simple device to measure the acceleration of gravity at a particular location. We measure the length l of the pendulum and then set the pendulum into motion. We measure the period by a clock and obtain the acceleration of gravity from equation 13.35 as
g = 4π^2 l (13.36) T p^2
Example 13.
The period of a pendulum. Find the period of a simple pendulum 1.50 m long.
Solution
The period, found from equation 13.35, is
T p = 2 ✜ (^) gl
= 2 ✜ 1.50 m 9.80 m/s 2 = 2.46 s
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Example 13.
The length of a pendulum. Find the length of a simple pendulum whose period is 1.00 s.
Solution
The length of the pendulum, found from equation 13.35, is
l = T p^2 g 4 π^2 = (1.00 s) 2 (9.80 m/s 2 ) 4 π^2 = 0.248 m
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Example 13.
The pendulum and the acceleration of gravity. A pendulum 1.50 m long is observed to have a period of 2.47 s at a certain location. Find the acceleration of gravity there.
Solution
The acceleration of gravity, found from equation 13.36, is
g = 4π^2 l T p^2 = 4 π^2 (1.50 m) (2.47 s) 2 = 9.71 m/s 2
To go to this Interactive Example click on this sentence.
We can also use a pendulum to measure an acceleration. If a pendulum is placed on board a rocket ship in interstellar space and the rocket ship is accelerated at 9.80 m/s^2 , the pendulum oscillates with the same period as it would at rest on the surface of the earth. An enclosed person or thing on the rocket ship could not distinguish between the acceleration of the rocket ship at 9.80 m/s^2 and the
