physics 110 Chapter 11: Vibration and waves PDF

Preview

Summary

Simple Harmonic Motion (SHM)
• Energy in Simple Harmonic Motion
• The Period and Sinusoidal Nature of SHM
• The Simple Pendulum
• Wave Motion
• Type of Waves: Transverse and Longitudinal • Energy Transported by Waves
• Reflection and Transmission of Waves
• ...

Description

Y. Abranyos (LaTex)

Chapter 11: Vibrations and Waves • Simple Harmonic Motion (SHM) • Energy in Simple Harmonic Motion • The Period and Sinusoidal Nature of SHM • The Simple Pendulum • Wave Motion • Type of Waves: Transverse and Longitudinal • Energy Transported by Waves • Reflection and Transmission of Waves • Interference; Principle of Superposition • Standing Waves and Resonance

1

Y. Abranyos (LaTex)

Simple Harmonic Motion Vibrational or oscillatory motions are one of the most common types in nature. Whenever a system in a stable equilibrium is slightly displaced from the equilibrium position it executes an oscillatory motion.

Unstable Equilibrium

Stable Equilibrium

We will study the following system: A mass m resting on a frictionless horizontal table and attached to a spring of spring constant k. When the spring is not compressed or stretched we have the system in equilibrium state and a displacement from equilibrium (compressing or stretching) it will initiate an oscillatory motion. Note that whenever we have a potential energy with a minimum it can lead to oscillation (see figure). The spring is stretched to maximum displacement amplitude A and released. Consequently, it will oscillate about the equilibrium position. PES(x) k

m

x

x −A

A (Amplitude)

x = 0 (Equilibrium Position)

Fs = −kx, Force exerted by the spring Fext = +kx, Force exerted by the external force

2

Y. Abranyos (LaTex) We want to study the oscillatory motion, and in particular we look for: • Given the position of the mass, what is its velocity? • Given the position of the mass, what is its acceleration? • What is the period of oscillation (time to move from A to −A and back to A)? • What are the position, velocity and acceleration as a function of time? We can always use newton’s 2nd Law to get the various quantities outlined. However, because the acceleration is not constant it is easier to use conservation of energy method.

11-2: Energy in Simple Harmonic Motion The total energy of the spring-mass system has KE due to the mass and P Es due to the spring. Since we have no friction the system is conservative and the total energy is constant. E = KE + P Es 1 1 E = mv 2 + kx2 2 2 1 2 1 2 1 2 1 2 kA = mv + kx = mv0 . 2 2 2 2 We see that the energy oscillates between maximum potential 21 kA2 which occurs when x = ±A, and maximum kinetic energy 12 mv02 which occurs at x = 0. The maximum velocity of the mass is at the equilibrium position x = 0 when the potential energy is zero. Setting x = 0 in the energy equation we obtain, 1 2 1 mv0 = kA2 2 2r k v0 = A . m 3

Y. Abranyos (LaTex) From the conservation of energy equation we can solve for the velocity as a function of position. Velocity as a function of position

r

k 2 (A − x2 ) k r  1/2 k x2 v = ±A 1− 2 A m   2 1/2 x . v = ±v0 1 − 2 A v = ±

The above formula can be used to obtain the velocity of the mass given its position. We have ± since we don’t know if it is moving to the right or to the left; we only know its magnitude. Acceleration as a function of position It is easier to use Newton’s 2nd Law to obtain the acceleration. The net force acting on the mass is only due to the spring.

Fnet = −kx = ma k a = − x. m

11-3 The Period and Sinusoidal Nature of SHM We now turn to the problem of determining the period of a mass m attached to a spring of spring constant k and executing an oscillatory motion with an amplitude A. To get the period of the system we use another type of motion, uniform circular motion with a velocity v0 with a radius of A the amplitude. See the following diagram. The red mass is moving at constant speed v0 in a circle of radius A and the blue mass is attached to a spring (not shown) and executes a SHM about 4

Y. Abranyos (LaTex)

ω A

θ x = −A

x=A x=0

x

P

x = 0 with an amplitude of A = r the radius of the circle . Both masses started at t = 0 at point P or θ = 0. For a uniform circular motion we have, v0 v0 = A r θ = ωt

ω =

The period of SHM We can see from the diagram above that the period of the red and blue masses are the same. The period of the red mass which is the same as the period of the blue mass is,

distance 2πr 2πA = = v0 v0 velocity r 2πA k q , ω= T = m A k T =

m

T = 2π

r

1 1 m and the frequency is, f = = k T 2π

r

k . m

We note here that the the period is independent of the amplitude. This is not in general true, it is true when the restoring force is proportional to the displacement, such as Hooke’s Law. 5

Y. Abranyos (LaTex) (11-5) A fisherman’s scale stretches 3.6 cm when a 2.4-kg fish hangs from it. (a) What is the spring stiffness constant and (b) what will be the amplitude and frequency of oscillation if the fish is pulled down 2.1 cm more and released so that it oscillates up and down? Solution: (a) The spring constant is found from the ratio of applied force to displacement. k=

mg (2.4)(9.80) F ext = = = 653 N/m x x 0.036

(b) The amplitude is the distance pulled down from equilibrium, so A = 2.1 cm. r r k 1 653 1 f= = 2.63 Hz = 2π m 2π 2.4 The Sinusoidal nature of SHM Sinusoidal means the function varies like a sine (cosine) function. The projection of the motion of the red mass on the x-axis coincides with the motion of the blue mass executing a SHM. Therefore, to learn about the SHM of the blue mass it suffice to study the x-motion of the red mass. As far as the x-motion is concerned, the red mass has the following motion given that, at t = 0, x(0) = A, v(0) = −ωA 2π T   t x = A cos(θ) = A cos 2π T θ = ωt, ω = 2πf =

  t v(t) = −ωA sin(ωt) = −ωA sin 2π T k a(t) = −ω 2 A cos(ωt) = −ω 2 x(t) = − x(t) m 6

Y. Abranyos (LaTex) Of course we could have started observing the two masses at a different time, such as t → t+T /4. This will change the cosine to sine. But, cosine and sine are essentially the same function the difference being if you displace cosine by 90o or π/2 radian to the right it becomes sine and the reverse. In general, we have x(t) = A cos(ωt + φ), φ is phase time that determines initial conditions v(t) = −ωA sin(ωt + φ) k a(t) = −ω 2 A cos(ωt + φ) = −ω 2 x(t) = − x(t) m Useful relationships: T π ⇔ 2 4 cos(θ ± 2π ) = ∓ sin(θ), sin(θ ± 2π) = ± cos(θ )

2πrad ⇔ 1period = T ⇒

Common initial conditions: 1. at t = 0 the mass is at x(0) = A x(t) = +A cos(ωt) 2. at t = 0 the mass is at x(0) = −A x(t) = −A cos(ωt) 3. at t = 0 the mass is at x(0) = 0 moving to the right v(0) = v0 = ωA x(t) = A sin(ωt) 4. at t = 0 the mass is at x(0) = 0 moving to the left v(0) = v0 = −ωA x(t) = −A sin(ωt) (11-8) A vertical spring with spring stiffness constant 305 N/m oscillates with an amplitude of 28.0cm when 0.235kg hangs from it. The mass passes through the equilibrium point (y = 0) with positive velocity at t = 0. 7

Y. Abranyos (LaTex) x(t) T A t −A cosine

sine

(a) What equation describes this motion as a function of time? (b) At what times will the spring be longest and shortest? Solution: We take downward as the positive direction of motion. r r 305 k = ω= rad/s = 36 rad/s m 0.235 (a) The mass has a zero displacement and a positive velocity at t = 0, the equation is a sine function. y(t) = A sin[ωt] y(t) = 0.280 sin[(36.0t] (b) The period of oscillation is given b, T = tmax:

tmin:

2π ω

=

2π 36

= 0.17441s.

y(tmax) = A ⇒ sin(ωtmax ) = 1 ⇒ ωtmax = 2πn + π/2 2πn + π/2 ⇒ tmax = = T (n + 1/4), n = 0, 1, 2, 3, . . . ω y(tmin ) = −A ⇒ sin(ωtmin ) = −1 ⇒ ωtmin = 2πn + 3π/2 2πn + 3π/2 ⇒ tmax = = T (n + 3/4), n = 0, 1, 2, 3, . . . ω

(11-20) An object with mass 2.7 kg is executing simple harmonic motion, attached to a spring with spring constant k = 310 N/m. When the object is 0.020 m from its equilibrium position, it is moving with a speed of 0.55 m/s. (a) Calculate the amplitude of the motion. 8

Y. Abranyos (LaTex) (b) Calculate the maximum speed attained by the object. Solution: (a) The total energy of an object in SHM is constant. r 1 2 1 2 1 2 m 2 E = mv + kx = kA ⇒ A = v + x2 2 2 2 k r 2.7 A= (0.552 ) + 0.022 = 5.51 × 10−2 m 310 (b) The energy is all kinetic energy when the object has its maximum velocity. r 1 1 2 1 2 k 2 E = mv + kx = kA ⇒ vmax = A m 2 2 2 r r 310 k −2 = 0.59 m/s = 5.51 × 10 vmax = A 2.7 m (11-22) Agent Arlene devised the following method of measuring the muzzle velocity of a rifle (Fig. 11-52). She fires a bullet into a 4.148-kg wooden block resting on a smooth surface, and attached to a spring of spring constant k = 162.7 N/m. The bullet, whose mass is 7.870 g, remains embedded in the wooden block. She measures the maximum distance that the block compresses the spring to be 9.460 cm. What is the speed v of the bullet? Solution: 1. Use conservation of linear momentum before and after collision (inelastic). 2. Use conservation of energy after collision. mv0 pbef ore = paf ter ⇒ mv0 = (m + M)v1 ⇒ v1 = m   +M 1 m 1 1 1 2 2 1 mv02 = kA2 Eaf ter = (m + M)v1 = kA ⇒ 2 2 2 2 m+M p p A 0.0946 ⇒ v0 = 162.7(0.0787 + 4.148) = 312.6 m/s k(m + M) = 0.00787 m 9

Y. Abranyos (LaTex) A bungee jumper with mass 65.0 kg jumps from a high bridge. After arriving at his lowest point, he oscillates up and down, reaching a low point seven more times in 43.0 s. He finally comes to rest 25.0 m below the level of the bridge. Estimate the spring stiffness constant and the unstretched length of the bungee cord assuming SHM. Solution: The period of the jumper’s motion is r 43 m 4π 2 m 4π 2 (65.0) T = = 68.0 N/m s = 6.14 s, T = 2π ⇒k= = 6.142 7 k T2 k∆x = mg ⇒ ∆x = 9.37 m ⇒ l0 = l − ∆x = (25.0 − 9.37) m = 15.6 m A 1.60-kg object oscillates at the end of a vertically hanging light spring once every 0.45 s. (a) Write down the equation giving its position y (± upward) as a function of time t. Assume the object started by being compressed 16 cm from the equilibrium position (where y = 0), and released. (b) How long will it take to get to the equilibrium position for the first time? (c) What will be its maximum speed? (d) What will be the object’s maximum acceleration, and where will it first be attained? (a) The mass, period, and amplitude are given. 2π = 14.0 rad/s T at t = 0 : y(0) = A, v = 0 ⇒ y(t) = A cos(ωt) = 0.16 cos(14t) A = 0.16 m, ω =

(b) The time to reach the equilibrium is one-quarter of a period

T 4

= 11.2.

(c) The maximum speed is given by vmax = ωA = 14.0 × 0.16 m/s = 2.2 m/s. (d) The maximum acceleration is ω 2 A = 14.02 × 0.16 m/s2 = 31 m/s2 The maximum acceleration occurs at the endpoints of the motion and is first attained at the release point. 10

Y. Abranyos (LaTex)

11-4 The Simple pendulum Another example of an oscillatory motion is the simple pendulum. It consists of a mass m attached to a massless rod or string and swings back and forth from a fixed point P (see diagram). We consider the case where the maximum opening angle θ0 (amplitude) is small. P

θ l FT x = lθ mg

θ mg sin(θ

The net force acting on the mass is mg sin(θ). Therefore,

Fnet = −mg sin(θ) = ma

x for small angles (in radians) we have sin(θ) ≈ θ = l x −mgθ = −mg = ma l If we compare this with Hooke′ s Law − kx = ma we see, g k = m . l Once we have the equivalent Hooke’s Law for the pendulum, everything we have done for the spring applies. In particular the period is given by, s r l m = 2π T = 2π k g r g 1 1 . = f = 2π l T 11

Y. Abranyos (LaTex) We see that the period does not depend on the mass and depends only on the length of the string and gravity. Example: What is the period of a simple pendulum 47 cm long (a) on the Earth, and (b) when it is in a freely falling elevator? s r l 0.47 T = 2π s = 1.4 s = 2π 9.8 g If the pendulum is in free fall, there is no acceleration ⇒ no period (T → ∞) Derive a formula for the maximum speed vmax of a simple pendulum bob in terms of g, the length l, and the maximum angle of swing θmax. Solution: Use conservation of energy Etop = Ebot ⇒ KEtop + P Etop = KEbot + P Ebot ⇒ P Etop = KEbot p p 1 2 mgh = mvmax ⇒ vmax = 2gh = 2gl(1 − cos θmax) 2

11-7 Wave Motion So far we have consider localized motion of object such as, particles, rigid bodies, fluids in a container or in motion. There are several physical phenomena that cannot be described in a localized model: waves. A wave is characterized as some sort of disturbance which propagates in space and time. There are several kinds of waves: mechanical waves (vibrating string or membrane), bulk wave propagating through a material (sound wave), seismic waves, electromagnetic waves propagating through a vacuum or materials (light) etc.. In general a propagating wave transports energy and momentum. Consider a long rope which is shaken at one end as shown in the diagram. The wave is not located at a specific point in space or time; rather it is spread out through space (1-dimensional in this case) and time. One cannot say the wave is located at a given point at a given time.

12

Y. Abranyos (LaTex)

Propagation direction

a

b

Vibration direction

11-8 Types of waves: Transverse and Longitudinal A slinky toy can be used to demonstrate two different kinds of waves. Shaken up and down at one end it excute a wave motion and we get a transverse wave. If on the other hand we perform compression and rarefaction at one end we get a longitudinal wave. Transverse waves In a transverse wave, the direction of motion of particles (vibration) is perpendicular to the direction of propagation of the wave. Examples: A vibrating string and electromagnetic waves. longitudinal waves In a Longitudinal wave, the direction of motion of particles (vibration) is parallel to the direction of propagation of the wave. Examples: Sound waves. Speed of a transverse waves on a string The speed of a mechanical wave depends on properties of the wave medium. Suppose we have a string of length L and mass m under a tension FT . It can be shown that the speed of the wave on the string to be 13

Y. Abranyos (LaTex)

v=

s

FT

=

s

FT m is the linear mass density. Here µ = L µ

m L

Example: A string of length 0.5 m has a mass of 125mg. The string is attached to a ceiling and a mass of 0.5 kg hangs from the other end. If it is shaken at the end at what speed does the transverse wave travel?

v= FT

s

FT m L

=

Mg = 0.5 × 9.8 = 4.9 N s 4.9 = 140 m/s. v= 1.25×10−4 0.5

Periodic waves A periodic wave repeats the same pattern over and over. The terminology for a periodic wave is similar to that used for uniform circular motion and simple harmonic motion. At any given point, the wave repeats itself after a time T the period. The inverse is the frequency, f=

1 T

After a time T , the wave has moved a distance λ called the wavelength since the wave repeats itself after every period. Therefore we have,

λ = vT λ v = = f λ. T 14

Y. Abranyos (LaTex) y(x,t = 0)

λ

A

λ x

−A

λ

The maximum displacement of any particle from its equilibrium position is the amplitude A of the wave. Harmonic waves are special kind of periodic waves in which the disturbance is sinusoidal (sine or cosine function). Example: A cord of mass 0.65 kg is stretched between two supports 8.0 m apart. If the tension in the cord is 120 N, how long will it take a pulse to travel from one support to the other? Solution: ∆x v= = ∆t

s

FT = µ

s

∆x FT = 0.21 s. ⇒ ∆t = q FT m/L m/L

11-9 Energy Transported by Waves For a sinusoidal wave of frequency f, the particles move in SHM as a wave passes, so each particle has an energy 1 E = kA2 2 Thus, we have the important result that the energy transported by a wave is proportional to the square of the amplitude. Just as with the oscillation that starts it, the energy transported by a wave is proportional to the square of the amplitude. From geometrical considerations, as long as the power output is constant, we see: Definition of intensity: 15

Y. Abranyos (LaTex)

Energy/time Power = area area Since the energy is proportional to the wave amplitude squared, so too is the intensity I=

I ∝ A2 If a wave flows out from the source in all directions, it is a three-dimensional wave. If a wave is able to spread out three-dimensionally from its source, and the medium is uniform, the wave is spherical (Ssphere = 4πr 2 ). P 1 , If we have constant power I ∝ 2 r 4πr 2 This is often called the inverse square law. I=

The amplitude of a wave also decreases with distance. A∝

1 r

Intensity Related to Amplitude and Frequency By looking at the energy of a particle of matter in the medium of the wave,we find: k = ω 2 m = 4π 2 f 2 m 1 2 kA = 2π 2 f 2 mA2 = 2π 2 f 2 ρV A2 = 2π 2 f 2 mA2 = 2π 2 f 2 ρSvtA2 2 E 1 P¯ = = 2π 2 f 2 mA2 = 2π 2 f 2 ρSvA2 = ρSvω 2 A2 2 t 1 2 2 for a string with µ ⇒ P¯ = µvω A 2

E=

16

Y. Abranyos (LaTex)

Example: A piano wire with mass 4.50 g and length. 120 cm is stretched with a tension of 30.0 N. A wave with frequency 115 Hz and amplitude 1.80 mm travels along the wire. (a) Calculate the average power carried by the wave. (b) What happens to the average power if the wave amplitude is halved? (a) We use v =

v=

s

q

FT , µ

ω = 2πf = 723rad/s, and µ =

FT =v= µ

s

m L

= 0.00375

30.0 = 90.0 m/s 0.0045/1.2

1 1 P¯ = µvω2 A2 = (3.75 × 10−3 )(90)(7232 )(0.00182 ) = 0.285 J/s = 0.285 W 2 2 (b) P¯ A P¯ ∝ A2 ⇒ A → ⇒ P¯ → 2 4

11-10 Reflection and Transmission of Waves Reflection A traveling wave transmits energy in the direction of propagation. At an abrupt boundary between one medium and another, reflection occurs. In sound we call this echo. For a traveling wave on a string, when the pulse 17

Y. Abranyos (LaTex) reaches the end where the sting is attached it will experience a force downward. The string pulls up on the wall and by Newton’s 3rd Law the wall pulls down the string and accelerates it downward. consequently we have a reflected wave propagating backward.

Incident

Reflected

Incident

Transmitted

11-11 Interference; Principle of Superposition Suppose two waves of the same type pass through the same region of space. When the amplitudes are not too large, the resulting net disturbance at any point the waves coincide is the sum of of the the individual disturbance due to each wave. This important principle is called The Principle of Linear Superposition. Interference The principle of superposition leads to very interesting effects when applied to coherent waves. Two waves are coherent if they have the same frequency and maintain a fixed phase relationship. Suppose two coherent waves pass through the same point in space. If the waves are in phase at that point, that is the the phase difference is a multiple of the wavelength they superpose constructively. If on the other hand the two waves in phase at that point, that is the the phase difference is a odd multiple of the half-wavelength they superpose destructively. 18

Y. Abranyos (LaTex)

Constructive interference The two waves are in phase

Destructive interference The two waves are out of phase

λ

λ λ /2