oscillations

Topic P–11–9

Oscillations

Contents 1 Periodic Motion 1.1 Period and frequency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Displacements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Describing periodic motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2 2 2 2

2 Simple Harmonic Motion 2.1 Equation of motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3 4

3 Spring–mass system

5

4 Energy consideration in simple harmonic motion

6

5 The 5.1 5.2 5.3

Pendulum Simple Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Physical Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Torsional Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6 Comparing simple harmonic motion & uniform circular motion

1

7 7 9 10 11

2

Periodic Motion

1

Periodic Motion

¤ Definition 1 (Periodic motion). Motions, processes or phenomena which repeat themselves are called periodic. The phenomena of periodic motions are so common, that its concepts are required for understanding physical effects in almost all branches of physics.

1.1

Period and frequency

¤ Definition 2. In a periodic motion, the smallest interval of time after which the process repeats itself is called its period. The reciprocal of the period is called the frequency of the process. Usually, the symbol T (SI units: second) is used to denote the period while ν (SI units: second−1 which is called hertz ; symbol Hz) is used to denote the frequency. Thus, we have ν=

1 . T

(1.1)

Also, ω (SI units: radian/second) is used to denote the circular frequency: ω=

1.2

2π = 2πν . T

(1.2)

Displacements

Motion represents change of some measurable physical quantity with time. We shall use the term displacement to collectively denote these quantities. Thus, displacements can be change is any physical quantity that can be measured such as position, angle, voltage, pressure or even electric and magnetic fields. In oscillatory motion, displacement variables are measured as functions of time. Displacements are generally measured with respect to some mean or equilibrium configuration. In mechanical dynamic systems, the equilibrium configuration is the configuration in which the net force on the system is zero. Displacement variables can take positive as well as negative values. In graphical representation of motion, the time variable is usually plotted along the horizontal axis and the displacement variable along the vertical axis. The next problem which arises is how to represent periodic motions mathematically. We study it next.

1.3

Describing periodic motion

Periodic motion can be represented only by periodic functions of time. A function f of t, f (t), is said to be periodic if there exists a real number T > 0 (called the period), such that f (t+T ) = f (t) for all t. This means that the function has the same value when the argument is increased or decreased by an integral multiple of the period. Simple example of periodic function are sine and cosine and their period is 2π radians. If the independent variable t is a dimensional quantity like time, we can use the periodicity of sine and cosine to construct periodic functions with an arbitrary period T : 2π t = sin ωt , T 2π g1 (t) = cos t = cos ωt . T

f1 (t) = sin

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3

Simple Harmonic Motion

Further, the functions with period T /n where n = 1, 2, 3, . . . also repeat their values after time T . Thus, we can construct two infinite sets of such periodic functions: 2nπ fn (t) = sin t = sin nωt , n = 1, 2, 3, . . . (1.3) T 2nπ t = cos nωt n = 0, 1, 2, 3, . . . gn (t) = cos (1.4) T In the set of cosine is included the constant function g0 (t) = 1 ,

(1.5)

which being constant is periodic for any value of T . The important theorem given by the French mathematician Jean Baptiste Joseph Fourier and which goes by his name asserts that an arbitrary (i.e. any) periodic function ψ(t) with period T can be represented by a unique linear combination of the functions fn (t) and gn (t). Mathematically, if the function ψ(t) has period T , then putting ω = 2π/T , we have ψ(t) = a0 + a1 cos ωt + a2 cos 2ωt + a3 cos 3ωt + · · · +b1 sin ωt + b2 sin 2ωt + b3 sin 3ωt + · · ·

(1.6)

where ai ’s and bi ’s are real constants, appropriately called Fourier coefficients. They can be uniquely determined by using mathematical methods called Fourier analysis. The sine and cosine functions are also called harmonic functions and since in periodic motion, the displacement can be expressed as the sum of these harmonic functions, periodic motion is also called harmonic motion. Finally, we note that from (1.6) that in a general periodic motion, harmonic terms with infinitely lots of frequencies appear (note the multiples of ω). The special periodic motion in whose representation, only one frequency appears, is called simple harmonic motion. We shall concentrate on this specific motion.

2


Since in mathematical representation of simple harmonic motion1 , only a single frequency appears, we can assume without any loss in generality, that the general displacement shall be of the form x(t) = a1 cos ωt + b1 sin ωt . (2.1) The right hand side of the above equation can be transformed as follows: Ã ! q a1 b1 2 2 a1 cos ωt + b1 sin ωt = a1 + b1 p 2 cos ωt + p 2 sin ωt . a1 + b21 a1 + b21 a1

b1 b1 and sin ϕ = − p 2 so that tan ϕ = − , we obtain 2 a1 + a1 + b1 q a1 cos ωt + b1 sin ωt = a21 + b21 (cos ϕ cos ωt − sin ϕ sin ωt) = A cos(ωt + ϕ), p where A = a21 + b21 . Thus, we see that, in simple harmonic motion, the displacement from some equilibrium position, can be always represented in the form: Put cos ϕ = p

a21

b21

x(t) = A cos(ωt + ϕ),

(2.2)

for two arbitrary constants A and ϕ. In this expression, A is called the amplitude of the motion which measures the maximum displacement from the equilibrium position; the argument of cosine is called the phase of the motion and thus ϕ represents the initial phase (i.e. the phase at t = 0) or the phase constant. 1 It is interesting to note that, in real life there are no simple harmonic motion, but a number of systems can be cited, whose oscillations are amazingly close to simple harmonic.

Anant Kumar



4


2.1

Equation of motion

Let us consider a mechanical dynamical system undergoing simple harmonic motion along the x–axis about some equilibrium position. Put the origin at the equilibrium position and let x denote its position coordinate. Then x also becomes its displacement. As such we must have x(t) = A cos(ωt + ϕ). Differentiating with respect to time, we obtain the velocity vx (t): vx (t) =

dx = −Aω sin(ωt + ϕ), dt

(2.3)

and differentiating once again, the acceleration ax : ax (t) =

d2 x = −Aω 2 cos(ωt + ϕ) = −ω 2 x. dt2

(2.4)

Thus, we see that the acceleration is proportional to the displacement and directed opposite to it (note the negative sign). This is the hallmark of simple harmonic motion. Thus, ¯ Theorem 1 (Simple Harmonic Motion). If in a dynamical system, the acceleration is proportional to the displacement from the equilibrium position and directed opposite to the displacement, the system will undergo simple harmonic motion. Its equation of motion will be of the form d2 x + ω 2 x = 0, dt2

(2.5)

x(t) = A cos(ωt + ϕ),

(2.6)

with solution for two arbitrary constants. Equation (2.5) is a differential equation and the solution (2.6) is known as the general solution of this differential equation because it represents the class of functions that will satisfy (2.5). To find out one particular solution, we must specify the values of A and ϕ. For this purpose, we require two initial conditions. The variation of the displacement x(t), velocity vx (t), and acceleration ax (t) as function of time is as shown in Figure 1. Note that the maximum displacement x(t) = A cos ωt vx (t) = −Aω cos ωt

ax (t) = −Aω 2 cos ωt

ω2 A

A T

ωA

2T

t T 2

3T 2

5T 2

3T

Figure 1: The variation of x(t), vx (t), ax (t) assuming ϕ = 0. is the amplitude A, the maximum speed is Aω attained at the equilibrium position while the maximum acceleration is Aω 2 attained when the displacement is maximum. Anant Kumar



5

Spring–mass system

3

Spring–mass system

Consider a physical system that consists of a block of mass m attached to the end of a massless ideal spring of spring constant k, with the block free to move on a horizontal, frictionless surface (see Figure 2). When the spring is neither compressed nor extended, there is no force on the block. This is the equilibrium configuration of the system. We put the origin of the coordinate system at the location of the block in this configuration and take the x axis parallel to the surface as shown. We know from experience that such a system oscillates back and forth if disturbed from its equilibrium position. x =0

The figure on the left shows a block attached to a spring moving on a frictionless surface. (a) When the block is at its equilibrium position (x = 0), the force exerted by the spring is zero. (b) When the block is displaced to the right of equilibrium (x > 0), the force exerted by the spring acts to the left. (c) When the block is displaced to the left of equilibrium (x < 0), the force exerted by the spring acts to the right.

x Fs = 0

(a)

Fs (b) Fs (c)

Figure 2: The simple harmonic oscillator.

We can understand the motion in Figure 2 by first recalling that when the block is displaced a small distance x from equilibrium, the spring exerts on the block a force whose magnitude is proportional to the displacement and given by Hooke’s law Fs = −kx.

(3.1)

This force is called restoring force because it is is always directed toward the equilibrium position and therefore opposite the displacement. Applying Newton’s second law to the block along the x axis, we obtain, Fs = −kx = max

⇒

ax = −

k x. m

That is, the acceleration of the block is proportional to the displacement from the equilibrium position and directed opposite to it. Therefore, this system of spring and mass will execute simple harmonic motion. By comparison with (2.4), we obtain that r k k 2 ω = ⇒ ω= . (3.2) m m The time period of oscillations of this spring mass system is therefore r 2π m T = = 2π . ω k

(3.3)

Now we can easily write the position of the block as a function of time: x(t) = A cos(ωt + ϕ). Anant Kumar



6

Energy consideration in simple harmonic motion

To determine A and ϕ, we need two initial conditions. Let us say, that at t = 0, the block was at a position xi and it was moving with a velocity vi . Then, we have the following two equations: x(t = 0) = xi = A cos ϕ,

and

vx (t = 0) = vi = −Aω sin ϕ.

Dividing the two equations, we obtain vi /xi = −ω tan ϕ or tan ϕ = −

vi . ωxi

(3.4)

Further, xi = A cos ϕ, and vi /ω = −A sin ϕ. Squaring and adding the two, we have ³ v ´2 i x2i + = A2 cos2 ϕ + A2 sin2 ϕ, ω from where we obtain

r A=

x2i +

³ v ´2 i

. (3.5) ω Thus the motion of the block is entirely found out. As special cases, we note the following two common situations: 1. The block is displaced to the right by an amount xm and released without any velocity. In this case, xi = xm and vi = 0, thus we obtain, ϕ = 0 and A = xm . Thus, for this case x(t) = xm cos ωt,

vx (t) = −ωxm sin ωt,

and ax (t) = −ω 2 xm cos ωt.

2. The block was released from the equilibrium position by imparting a velocity vm towards the right. In this case, xi = 0 and vi = vm . Thus, ϕ = −π/2, and A = vm /ω. Thus, in this case, vm sin ωt, vx (t) = vm cos ωt, and ax (t) = −ωvm sin ωt. x(t) = ω

4

Energy consideration in simple harmonic motion

Let us examine the mechanical energy of the block-spring system illustrated in the last section. Because the surface is frictionless, we expect the total mechanical energy to be constant. The kinetic energy of the system is the same as the kinetic energy of the block. Thus, K=

1 1 mvx2 = mω 2 A2 sin2 (ωt + ϕ), 2 2

(4.1)

and the potential energy is that of the spring. Thus, U=

1 2 1 kx = kA2 cos2 (ωt + ϕ). 2 2

(4.2)

We see that K and U are always positive quantities. Because ω 2 = k/m we can express the total mechanical energy of the simple harmonic oscillator as E =K +U =

¡ ¢ 1 kA2 sin2 (ωt + ϕ) + cos2 (ωt + ϕ) . 2

But since, sin2 θ + cos2 θ = 1, we obtain the total mechanical energy of the system as E=

1 kA2 . 2

(4.3)

That is, the total mechanical energy of a simple harmonic oscillator is a constant of the motion and is proportional to the square of the amplitude. Note that U is small Anant Kumar



7

The Pendulum

when K is large, and vice versa, because the sum must be constant. In fact, the total mechanical energy is equal to the maximum potential energy stored in the spring when x = ±A, because at these points vx = 0, and thus there is no kinetic energy. At the equilibrium position, where U = 0 because x = 0, the total energy, all in the form of kinetic energy, is again 12 kA2 . That is 1 1 1 k 1 2 mvmax = mω 2 A2 = m A2 = kA2 . (at x = 0) 2 2 2 m 2 Again transform (4.1) and (4.2) as follows: µ ¶ 1 1 1 2 2 2 2 1 − cos(2ωt + 2ϕ) K = mω A sin (ωt + ϕ) = kA = kA2 (1 − cos(2ωt + 2ϕ)) , 2 2 2 4 µ ¶ 1 1 1 + cos(2ωt + 2ϕ) 1 U = kA2 cos2 (ωt + ϕ) = kA2 = kA2 (1 + cos(2ωt + 2ϕ)) . 2 2 2 4 E=

From the above two expressions, we see that the kinetic and the potential energy oscillates with twice the frequency. Plots of the kinetic and potential energies versus time appear in Figure 3(a), where we have taken the phase constant ϕ = 0. As already mentioned, both K and U are always positive, and at all times their sum is a constant equal to 12 kA2 , the total energy of the system. Figure 3(b) shows the variation of the kinetic and potential energy as function of the position x. K, U

E=

1 2

K, U

kA2

T 4

T 2

3T 4

T

5T 2

3T 2

1 2

kA2

U

U

K

K t

0

E=

−A

0

x +A

(b)

(a)

Figure 3: (a) The variation of the kinetic energy K = 12 kA2 sin2 ωt, and the potential energy U = 12 kA2 cos2 ωt.

It is assumed that ϕ = 0. (b) Plot of K and U as function of x. In either plot, note that K + U is a constant equal to 12 kA2 .

Further, we can use the principle of conservation of energy to obtain the velocity for an arbitrary displacement by expressing the total energy at some arbitrary position x as 1 1 1 E = K + U = mv 2 + kx2 = kA2 2 2 2 r p k 2 ⇒ v =± (A − x2 ) = ±ω A2 − x2 (4.4) m Finally, Figure 4 illustrates the position, velocity, acceleration, kinetic energy, and potential energy of the block-spring system for one full period of the motion. Most of the ideas discussed so far are incorporated in this important figure. Study it carefully.

5 5.1

The Pendulum Simple Pendulum

The simple pendulum is another mechanical system that exhibits periodic motion. It consists of a particle–like bob of mass m suspended by a light string of length ` that is fixed at the upper Anant Kumar



8

The Pendulum t

x

v

a

K

U

0

A

0

– ω 2A

0

1 2 kA 2

T/4

0

– ωA

0

T/2

–A

0

ω 2A

0

3T/4

0

ωA ω

0

1 2 kA 2

0

T

A

0

– ω 2A

0

1 2 kA 2

a max

vmax

1 2 kA 2

0

amax

1 2 kA 2

vmax

a max

x –A

0

A

Figure 4: Simple harmonic motion for a spring–mass system. The parameters in the table refer to the spring–mass system, assuming that x = A at t = 0; thus x(t) = A cos ωt.

end, as shown in Figure 5. The motion occurs in the vertical plane and is driven by the force of gravity. We shall show that, provided the angle θ is small (less than about 10◦ ), the motion is simple harmonic. In this case, the position of the bob is uniquely determined by the angle θ that the string makes with the vertical. Take the positive direction of θ as shown in Figure 5. The forces acting on the bob are the force T~ exerted by the string and the gravitational force m~g . The tangential component of the gravitational force, mg sin θ, always acts toward θ = 0, opposite the displacement. Therefore, the tangential force is a restoring force. We apply the Newton’s second law in form of the torque equation: d2 θ τz = I 2 , dt where the torque is calculated about the point of suspension, the z axis passes through this point and is perpendicular to the plane, coming towards the reader (this direction of the z axis coincides with the positive θ direction in accordance with the right–hand thumb rule), while I is the moment of inertia about the z axis. The contribution to this torque is entirely because of the component mg sin θ of m~g as you can verify. Thus, −mg` sin θ = m`2 Anant Kumar

d2 θ dt2

⇒

d2 θ g = − sin θ, 2 dt `



9

The Pendulum

Pivot

O θ d

T~

`

θ CM ˆ n

mg sin θ

τˆ

m mg cos θ m~g

m~g

Figure 6: The physical pendulum.

Figure 5: The simple pendulum with all the forces (and its component) acting on the bob.

where the negative sign appears because this torque tends to rotate the pendulum in the direction of negative θ. This, then, is the equation governing the motion of a simple pendulum. The right side is proportional to sin θ rather than to θ; hence, with sin θ present, we would not expect simple harmonic motion because this expression is not of the form of Equation (2.5). However, if we assume that θ is small, we can use the approximation sin θ ≈ θ; thus the equation of motion for the simple pendulum becomes d2 θ g (5.1) + θ = 0. dt2 ` Comparison with (2.5), shows that we must have r g g 2 ω = ⇒ ω= . ` ` Thus the time period and the frequency of the simple pendulum becomes s r 2π 1 1 ` g T = = 2π , and ν= = . ω g T 2π `

(5.2)

Also, we obtain θ(t) = θmax cos(ωt + ϕ), where θmax is the maximum angular displacement of the pendulum bob from its equilibrium position. From (5.2), we note that the period and frequency of a simple pendulum depend only on the length of the string and the acceleration due to gravity. Because the period is independent of the mass, we conclude that all simple pendulums that are of equal length and are at the same location (so that g is constant) oscillate with the same period.

5.2

Physical Pendulum

An arbitrary rigid body supported at a point so as to oscillated freely constitutes a physical pendulum. Consider a rigid body pivoted at a point O that is a distance d from the center of mass (Figure 6). The force of gravity provides a torque about an axis through O, and the 2 magnitude of that torque is mgd sin θ, where θ is as shown in Figure 6. Using τz = I ddt2θ , where I is the moment of inertia about the axis through O, we obtain −mgd sin θ = I

Anant Kumar

d2 θ . dt2



10

The Pendulum

The minus sign indicates that the torque about O tends to decrease θ. That is, the force of gravity produces a restoring torque. If we again assume that θ is small, we can use the approximation sin θ ≈ θ, and the equation of motion reduces to µ ¶ d2 θ mgd =− θ = −ω 2 θ. (5.3) dt2 I Comparing with (2.5), we see that the motion of the physical pendulum is simple harmonic, with the time period s 2π I T = = 2π . (5.4) ω mgd Note that the above expression reduces to the time period of a simple pendulum if I = md2 , i.e. when all the mass is concentrated at the center of mass.

5.3

Torsional Pendulum

Figure 7 shows a rigid body suspended by a wire attached at the top to a fixed support. When the body is twisted through some small angle θ, the twisted wire exerts on the body a restoring torque that is proportional to the angular displacement. That is, τ = −κ θ, where κ (kappa) is called the torsion constant of the support wire. The value of κ can be obtained by applying a known torque to twist the wire through a measurable angle θ.

Balance wheel

O

θmax P

Figure 7: A torsional pendulum consists of

Figure 8: The balance wheel of this antique pocket

a rigid body suspended by a wire attached to a rigid support. The body oscillates about the line OP with an amplitude θmax .

watch is a torsional pendulum and regulates the time– keeping mechanism.

Applying Newton’s second law for rotational motion, we find τ = −κ θ = I

d2 θ dt2

κ d2 θ =− θ dt2 I

(5.5)

p Again, this is the equation of motion for a simple harmonic oscillator, with ω = κ/I and a period r I T = 2π . (5.6) κ This system is called a torsional pendulum. There is no small–angle restriction in this situation as long as the elastic limit of the wire is not exceeded. Anant Kumar



11

Comparing simple harmonic motion & uniform circular motion

6

Comparing simple harmonic motion & uniform circular motion

We can better understand and visualize many aspects of simple harmonic motion by studying its relationship to uniform circular motion. Figure 9 is an overhead view of an experimental arrangement that shows this relationship. A ball is attached to the rim of a turntable of radius A, which is illuminated from the side by a lamp. The ball casts a shadow on a screen. We find that as the turntable rotates with constant angular speed, the shadow of the ball moves back and forth in simple harmonic motion. Consider a particle located at point P on the circumference of a circle of radius A, as shown in Figure 10(a), with the line OP making an angle ϕ with the x axis at t = 0. We call this circle a reference circle for comparing simple harmonic motion and uniform circular motion, and we take the position of P at t = 0 as our reference position. If the particle moves along the circle with constant angular speed ω until OP makes an angle θ with the x axis, as illustrated in Figure 10(b), then at some time t > 0, the angle between OP and the x axis is θ = ωt + ϕ.

Lamp

Q Ball A

P Turntable Screen

A Shadow of ball

Figure 9: An experimental setup for demonstrating the connection between simple harmonic motion and uniform circular motion. As the ball rotates on the turntable with constant angular speed, its shadow on the screen moves back and forth in simple harmonic motion.

As the particle moves along the circle, the projection of P on the x axis, labeled point Q, moves back and forth along the x axis, between the limits x = ±A. Note that points P and Q always have the same x coordinate. From the right triangle OP Q, we see that this x coordinate is x = A cos(ωt + ϕ). (6.1) This expression shows that the point Q moves with simple harmonic motion along the x axis. Therefore, we conclude that simple harmonic motion along a straight line can be represented by the projection of uniform circular motion along a diameter of a reference circle. We can make a similar argument by noting from Figure 10(b) that the projection of P along the y axis also exhibits simple harmonic motion. Therefore, uniform circular motion can be considered a combination of two simple harmonic motions, one along the x axis and one along the y axis, with the two differing in phase by 90◦ . This geometric interpretation shows that the time for one complete revolution of the point P on the reference circle is equal to the period of motion T for simple harmonic motion between x = ±A. That is, the angular speed ω of P is the same as the angular frequency ω of simple harmonic motion along the x axis (this is why we use the same symbol). The phase constant ϕ for simple harmonic motion corresponds to the initial angle that OP makes with the x axis. The radius A of the reference circle equals the amplitude of the simple harmonic motion. Because the relationship between linear and angular speed for circular motion is v = rω, the particle moving on the reference circle of radius A has a velocity of magnitude ωA. From the geometry in Figure 10(c), we see that the x component of this velocity is −ωA sin(ωt + ϕ). By definition, the point Q has a velocity given by dx/dt. Differentiating Equation (6.1) with respect to time, we find that the velocity of Q is the same as the x component of the velocity of P . Anant Kumar



12

Comparing simple harmonic motion & uniform circular motion y

y

ω

P

A

A

P y

t=0 θ

ϕ x

O

O

x

Q

x

θ = ωt + ϕ

(b)

(a) y

y

~v

ax

P

vx

P ~a

O

vx

x

Q

O

ax

x

a = ω2 A

v = ωA

(c)

Q

(d)

Figure 10: Relationship between the uniform circular motion of a point P and the simple harmonic motion of a point Q. A particle at P moves in a circle of radius A with constant angular speed ω. (a) A reference circle showing the position of P at t = 0. (b) The x coordinates of points P and Q are equal and vary in time as x(t) = A cos(ωt + ϕ). (c) The x component of the velocity of P equals the velocity of Q. (d) The x component of the acceleration of P equals the acceleration of Q.

The acceleration of P on the reference circle is directed radially inward toward O and has a magnitude v 2 /A = ω 2 A. From the geometry in Figure 10(d), we see that the x component of this acceleration is −ω 2 A cos(ωt + ϕ). This value is also the acceleration of the projected point Q along the x axis, as you can verify by taking the second derivative of Equation (6.1).

Anant Kumar



oscillations

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