Academic Resource Center: Waves Workshop

Transcription

1 Academic Resource Center: Waves Workshop

2 Presentation Outline Understanding concepts Types of waves Transverse Waves Longitudinal Waves Interference Constructive v. Destructive Beating Practice problems Singing in the shower Tuning instruments by ear Caterpillar motion Interference

3 Wave Types Transverse Waves: Disturbance perpendicular to propagation Ocean waves caterpillar Longitudinal Waves: Disturbance parallel to propagation Traffic jams sound

4 Interference Constructive interference Destructive interference

5 Interference When two waves are similar in frequency or phase, but not exactly matching, they beat together.

6 Practice Problem: Singing in the Shower Many men like singing in the shower stall because somehow enhances their voice. How does this happen? Would the effect be different for men and women?* *Problem 17.8 Ohanian Physics for Engineers 3 rd Editions

7 Shower Singing: Solution Sound waves are trapped within the shower stall and bounce back and forth from the singer to the wall, as well as wall to wall. The sound waves interfere with themselves this way. If the distance between the wall and the singer (or wall-to-wall) is an halfinteger multiple of the wavelength, constructive interference creates standing waves which amplify the sound. Here the distance from singer to wall is 1.5*wavelength of the sound. The sound wave is amplified.

8 Shower Singing: Solution If the distance between the wall and the singer (or wall-to-wall) is not an halfinteger multiple of the wavelength, destructive interference occurs. Because women's voices are typically slightly higher than men's voices, the wavelengths of their singing would be slightly shorter than a half-integer multiple. This causes a beating phenomenon which sounds off-key.

9 Practice Problem: Tuning String instruments such as the guitar or piano need to be tuned often; the tension on the strings need to be adjusted so that each string plays a certain musical note. Expert players can tune strings by ear as they have before modern technology. One method involves playing an instrument that is already tuned and adjusting your instrument accordingly. Say you have one guitar that is properly tuned and another that is not. How can you tell when a particular string is tuned just right? (Hint: what would happen if the string was only slightly off-tune? What would it sound like?)

10 Tuning: Solution When you tune a string close to the right tension, the sound it would produce would have a wavelength close, but not exactly, to wavelength it should have. For example the low E string should produce a sound with wavelength 1670cm, but when it is slightly out of tune it produces 1650cm wavelength sounds. When the tuned string and out-of-tune string are plucked at the same time, the two waves, that have similar wavelengths, beat together. The sound of the beating would oscillate between loud and quiet until they both die down.

11 Practice Problem: Caterpillar A nature photographer took this picture of a caterpillar moving on a leaf. After taking the picture, he observed it to move at about 1 inch every 2 seconds. Estimate the wave number and frequency of the wave traveling through the caterpillar's body as it moves.

12 Caterpillar Problem: Solution The first quantity to be solved is wave number. The definition of wave number is 2π/λ. What is λ? It is the wavelength of the wave traveling through the caterpillar. Any wavelength can be measured from crest-to-crest,node-to-node, or troughto-trough. The picture can be used. From one end of the caterpillar to the other measures about 1 inch using the scale, approximately. Therefore 1 = λ; k = 6.3 in -1

13 Caterpillar Problem: Solution The second quantity to be solved is frequency. There are many relations involving frequency, e.g. ω = 2πf, however always look for the relationship that has what you are solving for (f) and what you already know (λ, k, v). You should notice you did not use one piece of information yet, the speed (1 inch per 2 seconds). This is a hint to find the relation between f and v. λf = v, or f = v/λ f = (1 /2sec)/(1 ) = 0.5sec-1 = 0.5 Hz.

14 Practice Problem: Interference Two transverse harmonic waves are described by y1 = Acos(πx - 3πt) and y2 = Acos(πx + 3πt) Where A = 5.0m, x is in meters, and t is in seconds. What is the maximum amplitude of the superposition of these two waves at x = 0.25m? What are the maximum transverse speed and acceleration at that point?* *Problem Ohanian Physics for Engineers 3 rd Editions

15 Interference Problem: Solution Setting x = 0.25cm, the super position of both waves can be described as y = 5cos(π/4 3πt) + 5cos(π/4 + 3πt) The two cosines can be combined using the identity cosθ 1 + cosθ 2 = 2cos((θ 1 -θ 2 )/2)*cos((θ 1 +θ 2 )/2) Therefore y = 5*2*cos(π/4+3πt)*cos(0) = 10cos(π/4 + 3πt) The maximum value of cosine is when the argument is zero. When it is zero, y = 10*cos(0) = 10 The maximum amplitude possible is 10m.

16 Interference Problem: Solution Speed and acceleration can be obtained by taking the first and second derivatives of the position y. y = 10cos(π/4 + 3πt) v = dy/dt = -30πsin(π/4 + 3πt) a = d 2 y/dt 2 = -90π 2 cos(π/4 + 3πt) The maximum speed is 30π m/s while the maximum acceleration is 90π 2 m/s 2. Note that these are maximum values; the actual speed and acceleration vary with time at the position x = 0.25m.