12. For each of the possible wave forms below, indicate which satisfy the wave equation, and which represent reasonable waveforms for actual waves on a string. For those which do represent waves, find the speed and direction of propagation, and sketch the waveform at t=0. The quantities a, b, c, A, and v are all positive constants. a. yx, t ax bt c (4 pts.) b. y x, t (4 pts.) x t a b (4 pts.) 2 1 ax b 2 c. y x, t A sin d. yx, t A sin ax 2 bt 2 (4 pts.) Solution a) y(x,t) = (ax + bt + c)2 This function can solve the wave equation, so in that sense it is a solution. However, it is a parabola which goes to infinity at large x, so it does not describe a reasonable wave. b) y(x,t) = 1/(ax2+b) This function has no time dependence, so cannot be a wave. c) y(x,t) = Asin(x/a + t/b) This can be rewritten y(x,t) = Asin( 1/a (x + at/b)). It is of the form f(x-vt) with v = -a/b. This is a reasonable wave form. It travels in the –x direction with speed a/b. d) y(x,t) = Asin(ax2 – bt2) This cannot be written in the form f(x-vt). Also, if you plug it in the wave equation it does not give a constant velocity at all points in space and time. Grader: 5 pts each. 3 for successfully plugging in, and 2 for explaining if it’s a good wave or not. (The first 2 points are for cases where partial credit is needed, some of these can be solve/explained without plugging into the wave equation). 13. (20 pts) It is observed that a pulse requires 0.1 second to travel from one end to the other of a long string. The tension in the string is provided by passing the string over a pulley to a weight which has 100 times the mass of the spring. (a) What is the length of the string? (b) What is the equation of the third normal mode? Solution 100 L g (a) L vt Plug in an expression for the velocity: v T where the tension is equal to the weight of the mass: 100Lg : L 100Lg t The mass density cancels: L 100Lg t Solve for L: L 100g t 2 L 9.8m Grader: 10 pts. They may plug in 0.1 seconds right away and have strange looking expressions with incorrect units, but that is OK if they get the right answer. (b) This is a clamped string for which we have written the normal modes before. Simply plug in 3 for n: 3x y An sin cos n t L Since we have some information on the velocity, the n can be given explicitly: 3x y An sin cosvkn t L The velocity was simplified above: 3x y An sin cos 10 Lg k n t L Use the standard definition of the wavenumber: 2 3x y An sin cos10 Lg n L t And the expression for the normal mode wavelengths for a clamped string: 2n 3x y An sin t cos10 Lg 2L L g 3x y An sin 3t cos10 L L g 3x y An sin t cos 30 L L For this problem, we found above that g and L are numerically equal, so you can express it like this: 3x y An sin cos30t L Grader: 10 pts. If they left it in terms of g and L to avoid an apparent problems with units, give full credit. 14. Consider a string extending form x = 0 to x = L as shown below. At x = L the string is fixed to a rigid wall, and at x = 0 the string is attached to a ring of negligible mass which is free to slide on a friction-free rod. (friction free, so no transverse force). x 0 L a) Using the appropriate boundary conditions, give the solution yn(x,t) for the nth normal mode, and give explicitly the frequency n and the wavelength n. (20 pts.) b) Sketch the patterns for the three lowest allowed modes. (10 pts.) Hints: Think carefully about the x=0 boundary condition. The size of the rod and massless ring are negligible. Perhaps a free body diagram would be in order! ----------------------------The right boundary condition is simply that for a clamped string: the y-value is zero at all times. yL, t 0 For the left boundary condition, consider a free body diagram of the massless ring: Frod Tstring The massless ring feels a horizontal force from the rod which opposes the horizontal component of the string tension. The lack of friction means that there will be no vertical force from the rod. Since the ring is massless, there must also be no vertical force from the string. If there were, the massless ring would instantly respond (with infinite acceleration) and move to a point where the vertical component of the tension is zero. Since the vertical tension is proportional to the slope of the string (dy/dx), the slope must be zero. So the boundary condition is: y 0 x x 0 Just as initial conditions can apply to the function (initial position) or its first derivative (initial velocity), boundary conditions can apply to the function (constant position) or its first derivative (constant slope). How you reach the answer depends on your initial guess for the string shape. Since the left side is zero and the origin is displaced, cosine would be a good choice: yx, t A coskx cost Apply the left boundary condition: y Ak sin cost 0 x x0 which tells us that = 0. Now apply the right boundary condition: yL, t A coskLcost 0 which is only true if: kn L km L n 2 2m 1 2 n 1,3,5,7,... m 1,2,3,4,5,.... Since k = 2/(always!), the normal mode wavelengths are: n 4L n n 1,3,5,7,... Since = v k(always!), the normal mode frequencies are: n nv 2L n 1,3,5,7,... The normal mode solutions are therefore: n nv y n A cos x cos t 2L 2L n 1,3,5,7,... Here are the patterns for the first three modes: Grader: There are many ways to do this problem! Correct identification and use of the two boundary conditions is worth 10 points. The boundary condition at x = L can only be as I stated it. For the boundary condition at x = 0, they may assume that the end will oscillate with amplitude A and get the same answer. Note that they may also neglect the phase if they choose wisely between sine and cosine to describe the shape (some will understand this, and some will get lucky). The expression for wavelengths is worth 5 pts, frequencies are worth 5 pts, and normal modes are worth 4pts. Note that these expressions may vary depending on choice of sine or cosine, and on choice of odd integers or all integers. The sketches are worth 2 pts each. 15. (20 pts) A stretched string of mass m, length L, and tension T is driven by two sources, one at each end. The sources both have the same frequency and amplitude A, but are exactly 180 degrees out of phase with respect to one another. (Each end is an antinode). What is the smallest frequency consistent with stationary vibrations of the string (a normal mode)? This is French problem 6-5. (20 pts.) Solution The stationary state with smallest frequency will be the first normal mode. Since both ends are anti-nodes, the n=1 normal mode has a wavelength equal to twice the string length L: A cost Acost 0 L 1 2L You can also get this result mathematically by guessing a solution: yn x, t An coskn x cosnt Its best to guess cosine since the string has amplitude at x = 0. Apply the left boundary condition: yn 0, t An cos0cosnt A cost This tells us that the amplitude of the normal mode (An) must just be the amplitude of the drive oscillation (A). Also, the frequency of the normal mode must be the frequency of the drive oscillation. Now for the right boundary condition: y n L, t A cosk n Lcost A cost Here I have plugged in L at the right boundary, and set it equal to the value of the left boundary, but 180 degrees out of phase. This is true if: cosk n L 1 which is true for k n L n odd n (1,3,5,7,9,...) Or you can handle the odd n this way: k n L 2n 1 Now to get the frequency: n v L n n all n (1,2,3,4,5,6,…). odd n (1,3,5,7,9,...) nv L The lowest frequency will be n=1: 1 v L In terms of the parameters given: 1 TL L m 1 T mL Grader: 10 pts for getting that the first normal mode wavelength is 2L (which can be done with equations or graphically), and 10 pts for finding the lowest frequency. If they stop too early and leave the expression for lowest frequency in terms of v (velocity) rather than T and m, take 2 pts.

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