viewnote9 - PHYS 302: Unit 9 Viewing Notes – Athabasca...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: PHYS 302: Unit 9 Viewing Notes – Athabasca University PHYS 302: Vibrations and Waves – Unit 9 Viewing Notes Getting to this point has been a lot of work! Now we’ll move on to some fun. Much of this lecture discusses sound and its generation. Moving objects can couple mechanical energy into the air, setting up pressure waves that are perceived as sound. Tuning forks do this (as already demonstrated), and a similar thing happens with stringed musical instruments. The basic equations for sound in an enclosure are similar to those for strings, but the boundary conditions are different. Starting with a review of strings is thus a good idea. In a string of length L with fixed ends and with a tension T and mass per unit length μ, disturbances propagate with the speed v T . From the previous lecture, the equation for the mode n in this case is yn An sin( n x ) cos(nt ) . The cos term represents time variation (i.e., the whole string moving up and down). L The sin term shows the shape of the string at any given time, a sin wave in x with an amplitude determined by the cos term’s value at time t and the overall amplitude An of that mode. For reasons that are not yet obvious, it is useful to express the standing wave parameters in terms of the wave speed. If λ is the wavelength, and f is the frequency in Hz (i.e., the number of waves per second), then λf waves go by any given point in one second. This must mean that the speed is v= λf. By meeting the boundary conditions we also found that an integral number of half-wavelengths must fit in the length L, so that 1 n n L , or n 2nL . With this wavelength, only one frequency is possible, 2 which is solved from v= λf. This is f n n nv 2L . For the fundamental, n=1, the wavelength is twice the string length (i.e., λ1=2L), and the frequency is f1 v 2L . For the first harmonic, n=2, a whole wavelength fits, so λ2=L and f2 v L v (3:00). Now Dr. Lewin considers different boundary conditions. If one end is free (which is hard to do with strings due to the difficulty of finding massless, frictionless strings) the boundary condition switches from requiring y to be zero at the end to requiring the slope to be zero. In this case, the fundamental has only one-quarter wave in it, so for n=1, 3 λ1=4L and f1 4vL . The next possible harmonic has ¾ of a wavelength, so for n=2, 2 4 L and f1 4 v . In L 3 general, n 24 L1 and f n n ( 2 n 1) v 4L . Instead of ratios 1,2,3,4… for fixed ends, here we have ratios 1,3,5,7…(7:00). A vibrating object is more efficient in coupling energy to the air if it is attached to a surface. In musical instruments, this surface is called a sounding board. For a stringed instrument with two fixed ends (i.e., all string instruments), the fundamental is f1 T 1 2L (9:40). If L is long, the pitch is lower, and if L is short, the pitch is higher. For higher tension or lower mass per unit length, one also gets a higher pitch. In a piano, the length and mass per unit length vary a lot, but the tension is about 200N for each string (about equivalent to supporting 2 kg on each string). The boundary condition of a wind instrument (18:10) depends on whether or not the end is open. If both ends of the instrument are open, then the ends are pressure nodes because they must have the same pressure as the air (the overpressure is 0). For our purposes, the speed of sound is 340 m/s, although Lewin mentions the general formula v RT M . With open ends, the formulas are the same as for a string, nv n 2nL and f n 2 L (19:30). By blowing air past the end, normal modes will get excited. A system that is open at one end and closed at the other is similar to the one-end free case, so f n (2 n 1) v 4L . A table is given (22:45) for open-open and open-closed systems (important for wind instruments). You should calculate these frequencies to verify them. Large systems, of course, have lower frequency. One can calculate the frequency for a particular open box. For the 256 Hz tuning fork, the box should be 33 cm long (and this is the case). With wind instruments, v is fixed, so the length is the only effective variable. Organ pipes come in various lengths; the holes in a flute allow the flautist to change the effective length depending on which hole(s) is/are covered. For a 16.6 cm open-open (ends) instrument, the frequency is about 1000 Hz. If the same instrument is closed at one end, the frequency is only about 512 Hz (demonstrated). Musical instruments often have many harmonics active at once, creating a complex wave pattern. A tuning fork is nearly in one mode only; 1 PHYS 302: Unit 9 Viewing Notes – Athabasca University a flute also has a very pure tone. Musical instruments showing more complex patterns are demonstrated until the break (48:40). The students play with instruments until class resumes (52:30). 1 2 y 2 y for transverse waves with ends v 2 t 2 x 2 separated by a distance L. The normal mode solutions are yn ( x, t ) An sin( n x ) cos(n t ) . Then, L vn n vkn L . For longitudinal waves, more boundary conditions are possible. If the system is open-open, then Dr. Lewin reminds us that we are solving the wave equation the solutions for overpressure p are the same as for displacement y in the transverse case. For open-closed systems, one gets k n (2 n 1) 4L instead, but still equation is n vkn . This can be generalized to two dimensions where the wave 1z z z , where z is the vertical displacement of the surface above the xy plane. The v 2 t 2 x 2 y 2 2 2 2 analog of fixed end boundary conditions is to fix all around the edges. This makes a vibrating square membrane, rather like a square drum. If the lengths are Lx and Ly, then the boundary conditions are that z=0 at x=0 and x=Lx, and z=0 at y=0 and y=Ly. It is easy to show that multiplying spatial parts similar to those for a string solves the wave equation and boundary conditions: z ( x, y , t ) Am , n sin m ,n vkm ,n v m Lx 2 n Ly sin cos m x Lx n y Ly m ,n t , where 2 (1:00:30). There are many possible normal modes. The lowest of these has n=m=1, and the whole membrane moves up and down. For m=2, n=1, there are two opposed regions in x, separated by a nodal line. On the nodal line the amplitude is 0. The next mode up would have four regions. This is illustrated well in the textbook. As a demonstration in real time, a plate is driven with powder on it. Where the nodes are, the plate is stationary and powder can simply sit in place. Such plates are called Chladni plates. The modes here are more complex than those of a membrane fixed at the edges. The modes change as the frequency is changed. We can now consider the three-dimensional wave equation. This applies to the case of a sound cavity of lengths a, b, c along the x, y, z directions with closed sides and open ends in the z direction. The wave equation is 1 2 p 2 p 2 p 2 p 2 2 2 , with solution pl ,m ,n Pl , m , n cos lax cos 2 2 v t x y z sin cos m y b n y c l ,m ,n t , where l=0,1,2,…, m=0,1,2,…, n=1,2,…. The condition on n is different from those on l and m, since n=0 would simply be nothing. Here, 0 ,0,1 v c l ,m ,n v la mb nc 2 2 2 . With c as the biggest dimension, the lowest mode is . Likely the next lowest modes also involve varying the last index. A demonstration with the speed of sound in helium (three times in the speed of sound in air) follows. Do not try this at home, please. 2 ...
View Full Document

Ask a homework question - tutors are online