# A: zero everywhere. B: positive everywhere. C: negative everywhere. D: depends on position.

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1 <ct > A string is clamped at both ends and then plucked so that it vibrates in a standing wave between two extreme positions a and c. (Let upward motion correspond to positive velocities.) When the string is in position b, the instantaneous velocity of points along the string is... <ct > A string is clamped at both ends and then plucked so that it vibrates in a standing wave between two extreme positions a and c. (Let upward motion correspond to positive velocities.) When the string is in position c, the instantaneous velocity of points along the string is A: zero everywhere. B: positive everywhere. C: negative everywhere. D: depends on position. Answer: Even though the string is flat at this point, some parts of the string will be moving up and other spots will be moving down. The velocity of any particular point depends on where it is located. A: zero everywhere B: positive everywhere. C: negative everywhere. D: depends on the position. Answer: Focus your attention on ANY particular spot on the string. That spot goes straight up and down. When it reaches its most extreme point, it is "turning around". At that instant, v=0 (you are "stopped" instantaneously.) This is true everywhere along the string. A is the correct answer here. At the extreme, the string is instantly at rest, every point is turning around <ct > Below is a picture of a standing wave on a 30 meter long string. What is the wavelength of the standing wave? <ct > Below is a picture of a standing wave on a 30 meter long string. What is the wavelength of the standing wave? 30 m A. 30 m B. 60 m C.15 m D. Impossible to tell Answer: The above picture shows ½ of a wavelength that takes up 30 m. Then, one wavelength is: 30m /( ½ ) = 60 m 30 m A. 30 m B) 60 m C) 10 m D) 20 m E) Impossible to tell Answer: The above picture shows 1 and ½ wavelengths that take up 30 m. Then, one wavelength is: 30m /(1 ½ ) = 20 m (There are other ways to think about this. E.g, notice that you can SEE one wavelength in the picture, just go 2/3 of the way across, I.e. 20 meters across ) 1

2 i Ct b ii A. Yes, n = 1 Ct Could you observe standing waves with a wavelength of 3 m on a string of length 2 m? (If so, what mode would it be in? ) B. Yes, n = 1.5 C.Yes, n = 1.33 Which of the two points on the string oscillates with the LARGER (higher) frequency? A) Left point (i) B) Right point (ii) C) They both have the same frequency Answer: When the string is up, both points are up, and when the string is down, both points are down. They are in lock step to one another so they have the same frequency (only their amplitudes are different) D.Yes (some other n.) E. No Answer: Standing waves must fit exactly with nodes at either end (i.e. n must be an integer) If the length of the string is 2m and the wavelength is 3m then: n = 2*Length / Wavelength = 2*2/3 = 1.33 Then, n is not an integer, so a standing wave cannot be formed. Here's another way to think about it: n=1 is a mode with wavelength 2L = 4 m (that's too big, it's not 3 m) n=2 is a mode with wavelength L = 2 m. (too small). Higher n => smaller wavelength. There is NO n with a wavelength of 3 m! A string vibrates with a fundamental frequency of 220 Hz. Besides 220 Hz, which of the following are "resonant frequencies" which you might also have? i) 110 Hz ii) 330 Hz iii) 440 Hz A: i only B: ii only C: iii only D: i and ii E: all three <ct > Answer: A string can have its harmonic and integer multiples of the harmonic, so if the harmonic is f 1 =220 Hz, then 2*f 1 =440Hz is alowed, but the others are not. CT b A string has a fundamental frequency f 1. The second harmonic has frequency f 2 = 2f 1, It s one octave higher! Which harmonic is TWO octaves above f 1? A) f 3 B) f 4 C) f 5 D) f 8 E)?? Answer: 2 octaves above the fundamental is 1 octave above the second harmonic: 2*f 2 = 2*(2*f 1 ) = 4*f 1 = f 4 2

3 How many interior nodes are there in a standing wave with n = 10? A. 9 B. 10 C. 11 <ct > D. I would have to draw it and count Answer: From the pictures we see that the number of node is 1 less then n. n=1 n=2 n=3 CT a A string vibrates in the fundamental, producing an A (440 Hz) sound. Suppose the speed of sound in the air could be suddenly doubled (but the string is left unchanged) What would you HEAR? A) Same pitch (440 Hz) B) Lower pitch C) Higher pitch, but not double D) Double pitch = one octave higher E)?? Answer: The frequency on a string is given by: f = n(v/2l) But v in this formula represents the speed of the wave ON THE STRING. It's independent of the speed of sound in air! (Don't mix them up, think about this!) So changing the v of sound in air has NO EFFECT - the string is STILL wiggling at 440 Hz, and you'll still HEAR 440 Hz! <ct b> v = T mass/length If the tension is increased by a factor of 9 (nine times the Tension force!) what! happens to the speed of waves on a string? A. Goes up by a factor of 3 B. Goes up by a factor of 4.5 C. Goes up by a factor of 9 D. Goes up by a factor of 81 E. None of these / I don t know Answer: The speed depends on the square root of tension, so a factor of 9 in tension corresponds to a factor of 3 in speed. What happens to the frequency of the v fundamental? f n = n " 2L Answer: A factor of 3 change in the speed will cause a factor of 3 change in the frequency.! <ct b> If you want to lower the pitch of a guitar string by two octaves, what must be done to its tension? A. Raise it by a factor of 4 B. Lower it by a factor of 4 C. Lower it by a factor of 2 D. Lower it by a factor of 16 E. None of these / I don t know Answer: The speed and frequency are proportional to the square root of the tension. To lower the frequency by 2 octaves ( 1 / 4 the frequency) the speed must be quartered, so the tension must be decreased by a factor of 4 2 =16. This is not a very realistic way to change the frequency of a guitar, the string would be way too floppy. 3

4 CT a The clothesline is being driven (frequency f ) and is in the 2nd harmonic (one node in the middle) Now I ADD some weights on the end, increasing T a little. What happens? A) Looks same, wiggles faster B) A second node appears C) Goes into the fundamental D) Lose the resonance (just flops a little, not pretty ) E)?? Answer: Increasing the tension will increase the harmonic frequencies. If the change is small, then the string will no longer be being driven at a resonant frequency, so it will just flop around. CT c If you increase Tension by a factor of 4 (so the speed doubles) A) The frequency of the fundamental doubles, all other harmonics stay the same as they were B) The frequency of EVERY harmonic doubles C) None of the frequencies change, the wavelengths double D) f 1 goes up by 2, f 2 by 4, (etc ) E) Something else/??? Answer: Quadrupling the tension will double the speed and therefore double ALL of the harmonics because they are given by: v f n = n " 2L! put your finger down one half of the way along the string, and then pluck, you are mostly likely to hear A: Still 440 Hz B: 220 Hz C: 880 Hz ct a D: Something entirely different Answer: Putting your finger down half way will force a node to be there. This will eliminate all of the odd harmonics of the string. You will hear, primarily, the second harmonic of the string, 880Hz. ct b put your finger down lightly, one third of the way along the string, and then pluck the longer side, you are mostly likely to hear A: 3*440Hz B: (1/3)*440 Hz C: 3/2 * 440 Hz D: 2/3 * 440 Hz E: Something entirely different Answer: Putting your finger down 1 / 3 of the way will force a node to be there. This will eliminate all of the harmonics that do not have this node. You will hear, primarily, the third harmonic of the string, 3*440Hz = 1320Hz. This is different then clamping the string 1 / 3 of the way!! (See lecture notes if you're confused, make sense of this for yourself!) 4

5 ct b2 put your finger down hard (pushing the string to the fret), one third of the way along the string, and then pluck the longer side, you are mostly likely to hear A: 3*440Hz B: (1/3)*440 Hz C: 3/2 * 440 Hz D: 2/3 * 440 Hz E: Something entirely different Answer: Putting your finger down hard 1 / 3 of the way will effectively clamp the string and shorten its length. When you play it now it is essentially 2 / 3 the original length, so you will likely hear 440Hz/( 2 / 3 ) = 3 / 2 *440Hz c Electric guitars can have several pickups to detect string motion. You can flip a switch to activate single pickups or combos. Which pickup would you activate to make your music sound crisper (i.e. pick up more of the higher pitch frequencies?) A. B. C. D. Some combo / no diff Answer: Low frequencies vibrate primarily in the middle of the string (look at the picture of the fundamental, particularly!) whereas high frequencies can vibrate all along the string. The pickups at A are better suited for these high frequencies. 5

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