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Instructions: Here are some practice problems to make you familiar with these concepts. Please review the following. 1. When properly tuned, the 6 strings of a guitar at their full length of L = 0.650 m are intended to produce the following frequencies: String Note Frequency Wave speed (Hz) (m/s) 1 82.4 E2 A2 2 110.0 3 D3 146.8 G3 B3 E4 4 196.0 5 246.9 329.6 Use the frequency equation to calculate the wave speed v in each string. 2. Sav that string 2 is improperly tuned, so that its wave speed is 11 m/s faster than the id

Instructions: Here are some practice problems to make you familiar with these concepts. Please review the following. 1. When properly tuned, the 6 strings of a guitar at their full length of L = 0.650 m are intended to produce the following frequencies: String Note Frequency Wave speed (Hz) (m/s) 1 82.4 E2 A2 2 110.0 3 D3 146.8 G3 B3 E4 4 196.0 5 246.9 329.6 Use the frequency equation to calculate the wave speed v in each string. 2. Sav that string 2 is improperly tuned, so that its wave speed is 11 m/s faster than the id

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Instructions: Here are some practice problems to make you familiar with these concepts. Please review the following. 1. When properly tuned, the 6 strings of a guitar at their fullI length of L = 0.650 m are intended to produce the following frequencies: Wave speed (m/s) String Note Frequency (Hz) 1 82.4 E2 A2 2 110.0 3. D3 146.8 4 G3 196.0 B3 246.9 6. E4 329.6 Use the frequency equation to calculate the wave speed v in each string. 2. Say that string 2 is improperly tuned, so that its wave speed is 11 m/s faster than the ideal speed you calculated in Question 1. What frequency would it produce? Is that frequency too high (“sharp") or too low (“flat")? By how many Hz is the note out of tune? 3. If string 4 is improperly tuned, so that it plays 5.2 Hz too low, what is the actual wave speed in the string? Do you need to increase or decrease the speed to get the note in tune? By how much must v change? Now let's consider the placement of the frets. We will use String 3, though the same pattern works for all strings. At its full length of L = 0.650 m, String 3 produces a frequency of 146.8 Hz (a D). To play other notes, we need to put frets at the right positions to produces notes in half-step increments (D#, E, F, etc.). There are the first few higher frequencies we want String 3 to produce.
Transcribed Image Text:Instructions: Here are some practice problems to make you familiar with these concepts. Please review the following. 1. When properly tuned, the 6 strings of a guitar at their fullI length of L = 0.650 m are intended to produce the following frequencies: Wave speed (m/s) String Note Frequency (Hz) 1 82.4 E2 A2 2 110.0 3. D3 146.8 4 G3 196.0 B3 246.9 6. E4 329.6 Use the frequency equation to calculate the wave speed v in each string. 2. Say that string 2 is improperly tuned, so that its wave speed is 11 m/s faster than the ideal speed you calculated in Question 1. What frequency would it produce? Is that frequency too high (“sharp") or too low (“flat")? By how many Hz is the note out of tune? 3. If string 4 is improperly tuned, so that it plays 5.2 Hz too low, what is the actual wave speed in the string? Do you need to increase or decrease the speed to get the note in tune? By how much must v change? Now let's consider the placement of the frets. We will use String 3, though the same pattern works for all strings. At its full length of L = 0.650 m, String 3 produces a frequency of 146.8 Hz (a D). To play other notes, we need to put frets at the right positions to produces notes in half-step increments (D#, E, F, etc.). There are the first few higher frequencies we want String 3 to produce.
For string 1 f=82.4 Hz , L=0.65 mf=v2Lv=2fLv=2×82.4×0.65v=107.12 m/s For string 2 v=2x110x0.65v=143 m/s For string 3 v=2×146.8×0.65v=190.84 m/s For string 4 v=2x196x0.65v=254.8 m/s For string 5 v=246.9×2x0.65v=320.97 m/s For string 6 v=2×329.6×0.65v=428.48 m/s
Transcribed Image Text:For string 1 f=82.4 Hz , L=0.65 mf=v2Lv=2fLv=2×82.4×0.65v=107.12 m/s For string 2 v=2x110x0.65v=143 m/s For string 3 v=2×146.8×0.65v=190.84 m/s For string 4 v=2x196x0.65v=254.8 m/s For string 5 v=246.9×2x0.65v=320.97 m/s For string 6 v=2×329.6×0.65v=428.48 m/s
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