Sine Waves: Transverse Speed, Acceleration, and Energy Transfer, Schemes and Mind Maps of Particle Physics

Download Sine Waves: Transverse Speed, Acceleration, and Energy Transfer and more Schemes and Mind Maps Particle Physics in PDF only on Docsity! Waves Transverse Speed and Acceleration Power of a Wave Lana Sheridan De Anza College May 21, 2020 Last time • solutions to the wave equation • sine waves (covered in lab) Question Quick Quiz 16.21 A sinusoidal wave of frequency f is traveling along a stretched string. The string is brought to rest, and a second traveling wave of frequency 2f is established on the string. What is the amplitude of the second wave? (A) twice that of the first wave (B) half that of the first wave (C) the same as that of the first wave (D) impossible to determine 1Serway & Jewett, page 489. Overview • transverse speed of an element of the medium • energy transfer by a sine wave Sine waves Consider a point, P, on a string carrying a sine wave. Suppose that point is at a fixed horizontal position x = 5λ/4, a constant. The y coordinate of P varies as: y ( 5λ 4 , t ) = A sin(−ωt + 5π/2) = A cos(ωt) The point is in simple harmonic motion! 490 Chapter 16 Wave Motion Substitute A 5 15.0 cm, y 5 15.0 cm, x 5 0, and t 5 0 into Equation 16.13: 15.0 5 115.0 2 sin f S sin f 5 1 S f 5 p 2 rad Write the wave function: y 5 A sin akx 2 vt 1 p 2 b 5 A cos 1kx 2 vt 2 (B) Determine the phase constant f and write a general expression for the wave function. S O L U T I O N Substitute the values for A, k, and v in SI units into this expression: y 5 0.150 cos (15.7x 2 50.3t) Sinusoidal Waves on Strings In Figure 16.1, we demonstrated how to create a pulse by jerking a taut string up and down once. To create a series of such pulses—a wave—let’s replace the hand with an oscillating blade vibrating in simple harmonic motion. Figure 16.10 repre- sents snapshots of the wave created in this way at intervals of T/4. Because the end of the blade oscillates in simple harmonic motion, each element of the string, such as that at P, also oscillates vertically with simple harmonic motion. Therefore, every element of the string can be treated as a simple harmonic oscillator vibrating with a frequency equal to the frequency of oscillation of the blade.2 Notice that while each element oscillates in the y direction, the wave travels to the right in the 1x direction with a speed v. Of course, that is the definition of a transverse wave. If we define t 5 0 as the time for which the configuration of the string is as shown in Figure 16.10a, the wave function can be written as y 5 A sin (kx 2 vt) We can use this expression to describe the motion of any element of the string. An ele- ment at point P (or any other element of the string) moves only vertically, and so its x coordinate remains constant. Therefore, the transverse speed vy (not to be confused with the wave speed v) and the transverse acceleration ay of elements of the string are vy 5 dy dt d x5constant 5 'y 't 5 2vA cos 1kx 2 vt 2 (16.14) ay 5 dvy dt d x5constant 5 'vy 't 5 2v2 A sin 1kx 2 vt 2 (16.15) These expressions incorporate partial derivatives because y depends on both x and t. In the operation 'y/'t, for example, we take a derivative with respect to t while holding x constant. The maximum magnitudes of the transverse speed and trans- verse acceleration are simply the absolute values of the coefficients of the cosine and sine functions: vy , max 5 vA (16.16) ay , max 5 v2A (16.17) The transverse speed and transverse acceleration of elements of the string do not reach their maximum values simultaneously. The transverse speed reaches its max- imum value (vA) when y 5 0, whereas the magnitude of the transverse acceleration 2In this arrangement, we are assuming that a string element always oscillates in a vertical line. The tension in the string would vary if an element were allowed to move sideways. Such motion would make the analysis very complex. P t = 0 t = T A P P P l 4 1 t = T 2 1 t = T 4 3 a b c d x y Figure 16.10 One method for producing a sinusoidal wave on a string. The left end of the string is connected to a blade that is set into oscillation. Every element of the string, such as that at point P, oscillates with simple harmonic motion in the vertical direction. ▸ 16.2 c o n t i n u e d Finalize Review the results carefully and make sure you understand them. How would the graph in Figure 16.9 change if the phase angle were zero? How would the graph change if the amplitude were 30.0 cm? How would the graph change if the wavelength were 10.0 cm? Sine waves: Transverse Speed and Transverse Acceleration vy = −ωA cos(kx −ωt) ay = −ω2A sin(kx −ωt) = −ω2y If we fix x =const. these are exactly the equations we had for SHM! The maximum transverse speed of a point P on the string is when it passes through its equilibrium position. vy ,max = ωA The maximum magnitude of acceleration occurs when y = A (or max value, including sign when y = −A). ay = ω2A Questions Can a wave on a string move with a wave speed that is greater than the maximum transverse speed vy ,max of an element of the string? (A) yes (B) no Questions Can the wave speed be much greater than the maximum element speed? (A) yes (B) no Sine waves: Transverse Speed and Transverse Acceleration vy = −ωA cos(kx −ωt) ay = −ω2A sin(kx −ωt) = −ω2y Rate of Energy Transfer in Sine Wave Waves do transmit energy. A wave pulse causes the mass at each point of the string to displace from its equilibrium point. At what rate does this transfer happen? (Find dE dt ) Consider the kinetic and potential energies in a small length of string. Kinetic: dK = 1 2 (dm)v2y Replacing vy : dK = 1 2 (dm)A2ω2 cos2(kx −ωt) Rate of Energy Transfer in Sine Wave Waves do transmit energy. A wave pulse causes the mass at each point of the string to displace from its equilibrium point. At what rate does this transfer happen? (Find dE dt ) Consider the kinetic and potential energies in a small length of string. Kinetic: dK = 1 2 (dm)v2y Replacing vy : dK = 1 2 (dm)A2ω2 cos2(kx −ωt) Rate of Energy Transfer in Sine Wave dK = 1 2 µ dxA2ω2 cos2(kx −ωt) dU = 1 2 µA2ω2 cos2(kx −ωt) dx Adding dU+ dK gives dE = µω2A2 cos2(kx −ωt) dx Integrating over one wavelength gives the energy per wavelength: Eλ = µω2A2 ∫λ 0 cos2(kx −ωt) dx = µω2A2λ 2 Rate of Energy Transfer in Sine Wave For one wavelength: Eλ = 1 2 µω2A2λ Power averaged over one wavelength: P = Eλ T = 1 2 µω2A2 λ T Average power of a wave on a string: P = 1 2 µω2A2v Question Quick Quiz 16.52 Which of the following, taken by itself, would be most effective in increasing the rate at which energy is transferred by a wave traveling along a string? (A) reducing the linear mass density of the string by one half (B) doubling the wavelength of the wave (C) doubling the tension in the string (D) doubling the amplitude of the wave 2Serway & Jewett, page 496.