Physics 18 Spring 2011 Homework 13 - Solutions Wednesday April 20, 2011

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1 Physics 18 Spring 011 Homework 13 - s Wednesday April 0, 011 Make sure your name is on your homework, and please box your final answer. Because we will be giing partial credit, be sure to attempt all the problems, een if you don t finish them. The homework is due at the beginning of class on Wednesday, April 7th. Because the solutions will be posted immediately after class, no late homeworks can be accepted! You are welcome to ask questions during the discussion session or during office hours. 1. A string fixed at both ends resonates at a fundamental frequency of 180 Hz. Which of the following will reduce the fundamental frequency to 90 Hz? (a) Double the tension and double the length. (b) Hale the tension and keep the length and the mass per unit length fixed. (c) Keep the tension and the mass per unit length fixed and double the length. (d) Keep the tension and the mass per unit length fixed and hale the length. The frequency is gien in terms of the waelength and elocity as f = /λ. Now the speed of the wae depends on the tension and mass per unit length as = T/µ, and so f = 1 T λ µ. Now, the fundamental frequency of the wae has a waelength of L, where L is the length of the string. So, f 1 = 1 T L µ. If we want to make the frequency half of what it was, f 1 f 1, then we should follow method (c), where we keep the tension and mass per unit length fixed, and double the length, which will gie ( ) f = 1 T 4L µ = 1 1 T = f 1 L µ. 1

2 . Estimate the resonant frequencies that are in the audible range of human hearing of the human ear canal. Treat the canal as an air column open at one end, stopped at the other end, and with a length of 1.00 in. How many resonant frequencies lie in this range? Human hearing has been found experimentally to be the most sensitie at frequencies of about 3, 9, and 15 khz. How do these frequencies compare to your calculations? The ear is an open-closed tube, and so has allowed standing-wae frequencies of f n = n = f 4L 1n, where n = 1, 3, 5,, and f 1 = = 343 = 3376 Hz, or khz, is 4L the fundamental frequency (we hae recalled that one inch is.54 centimeters. The next few frequencies are f 3 = 3f 1 = 3(3.376) = khz f 5 = 5f 1 = 5(3.376) = 16.9 khz f 7 = 7f 1 = 7(3.376) = 3.6 khz. Now, the high-end of audible frequencies for the human ear is about 0 khz, so the n = 7 mode isn t within range. So, there are only three resonant modes, n = 1, 3, 5, and the theoretical alues are fairly close to the experimental alues.

3 3. It is thought that the brain determines the direction of the source of a sound by sensing the phase difference between the sound waes striking the eardrums. A distance source emits sound of frequency 680 Hz. When you are directly facing a sound source there is no phase difference. Estimate the phase difference between the sounds receied by your ears when you are facing 90 away from the direction of the source. The phase difference is just due to the extra distance that one wae traels oer another, δ = π x. We can rewrite this in terms of the speed and frequency of the wae, = λf, λ giing δ = π xf. Suppose that there is a distance of about 0 centimeters between your ears. Then, since the speed of sound is 343 m/s, we hae δ = πf x = π (680). 343 =.49 rad. 3

4 4. A standing wae on a rope is represented by the wae function ( ) 1 y(x, t) = (0.00) sin πx cos(40πt), where x and y are in meters, and t is in seconds. (a) Write wae functions for two traeling waes that, when superimposed, produce this standing-wae pattern. (b) What is the distance between the nodes of the standing wae? (c) What is the maximum speed of the rope at x = 1.0 m? (d) What is the maximum acceleration of the rope at x = 1.0 m? (a) Recall that the sum of two sine waes is [ ] [ ] 1 1 sin A + sin B = cos (A B) sin (A + B). This is precisely the correct form of the gien standing wae, so we can write ( π ) ( π ) y (x, t) = sin x 40πt sin x + 40πt, which gies the superposition of a wae traeling to the right and another traeling to the left. (b) The distance between the nodes is just half the waelength. Now, the wae ector k = π/λ = π/, so λ = 4 meters, and thus the distance between the nodes is d = meters. (c) The speed of the rope (not the wae) is just the deriatie of the displacement, = ẏ. Taking the deriatie gies ( ( = d dt (0.00) sin 1 πx) cos(40πt) ) = (0.00) sin ( 1 πx) d [cos(40πt)] dt = (0.800)π sin ( 1 πx) sin(40πt). Now, the maximum speed occurs when the cosine is 1 (to cancel the minus sign in the elocity). So, the maximum elocity is = 0.8π sin ( π x). When x = 1, then ( π ) (x = 1) = 0.8π sin = 0.8π m/s =.5 m/s. (d) The acceleration is just the second deriatie of the position, a = ÿ, or ( ( a = d dt (0.00) sin 1 πx) cos(40πt) ) = (0.00) sin ( 1 πx) d [cos(40πt)] dt = 3π sin ( 1 πx) sin(40πt). Again, the maximum acceleration occurs when the cosine is 1, and when x = 1, we find a max = 3π m/s. 4

5 5. Ultrasound has many medical applications, one of which is to monitor fetal heartbeats by reflecting ultrasound off a fetus in the womb. (a) Consider an object moing at speed 0 toward an at-rest source that is emitting sound waes of frequency f 0. Show that the reflected wae (i.e., the echo) that returns to the source has a Doppler-shifted frequency ( ) + 0 f echo = f 0, 0 where is the speed of sound in the medium. (b) Suppose that the object s speed is much less than the wae speed: 0. then f echo f 0, and a microphone that is sensitie to these frequencies will detect a beat frequency if it listens to f 0 and f echo simultaneously. Use the binomial expansion and other appropriate approximations to show that the beat frequency is f beat 0 f 0. (c) The reflection of.40 MHz ultrasound waes from the surface of a fetus s beating heart is combined with the.40 MHz wae to produce a beat frequency that reaches a maximum of 65 Hz. What is the maximum speed of the surface of the heart? The speed of ultrasound waes within the body is 1540 m/s. (d) Suppose the surface of the heart moes in simple harmonic motion at 90 beats/min. What is the amplitude in mm of the heartbeat? (a) The source is at rest and is sending out the waes, which leads to a (higher) Doppler-shifted frequency receied by the object, f R = f 0 (1 + 0 /). This frequency is reflected back to the source, but the object now looks like a moing source, and the frequency is Doppler-shifted for a second time, f echo = Plugging in for f R from before gies ( ) ( ) / + 0 f echo = f 0 = f 0, 1 0 / 0 as claimed. f R 1 0 /. (b) The beat frequency is just the difference between the echo frequency and the originalo frequency, f echo f 0. Now, recalling that 0, then, recalling that (1 + ɛ) α 1 + αɛ, for small ɛ, ( ) f echo = 1+0 / 1 0 f / 0 = ( ) ( ) 1 f0 ( ) ( ) ( ) f f0, 5

6 where we hae dropped terms of order ( 0 /) and higher. Thus, ( f beat f echo f 0 = 1 + ) 0 f 0 f 0 = 0 f 0. (c) Here we just plug in the numbers to find 0 using our results from part (b), or about.1 cm/s. 0 = f beat f 0 = = m/s, (d) The amplitude, A, is gien in terms of the maximum elocity as max = 0 = ωa, where ω = πf is the angular frequency. So, which is about. mm. A = 0 πf = π (90/60) = 0.00 m, 6