(7%) Problem 6: Using special techniques called string harmonics (or-flageolet t
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(7%) Problem 6: Using special techniques called string harmonics (or-flageolet tones ), stringed instruments can produce the first few overtones of the harmonic series. While a violinist is playing some of these harmonics for us, we take a picture of the vibrating string (see figures). Using an oscilliscope, we find the violinist plays a note with frequency f 805 Hz in figure (a). Otheexpertta.com 14% Part (a) How many nodes does the standing wave in figure (a) have? ia 14% Part (b) How many antinodes does the standing wave in figure (a) have? i 14% Part (c) The string length of a violin is about L 33 cm. What is the wavelength of the standing wave in figure (a) in meters? 14% Part (d) The fundamental frequency is the lowest frequency that a string can vibrate at (see figure b)). What is the fundamental frequency for our violin in Hz? 14% Part (e) In terms of the fundamental frequency what is the frequency ofthe note the violinist is playing in figure (c)? 14% Part (f) Write a general expression for the frequency of any note the violinist can play in this manner, in terms of the fundamental frequency fi and the number of antinodes on the standing wave A 14% Part (g) What is the frequency, in hertz, of the note the violinist is playing in figure (d)? Grade Summary Deductions 0%Explanation / Answer
A). The nodes are the point on the string that never move at all i.e the displacement of the points is zero for all time. There are 5 such points for the wave in figure a).
B) The points midway between the nodes care called antinodes. There are 4 antinodes for wave shown in figure a).
C) The lengths of the wave is L = 33cm. In figure a) There are 2 wavelengths i.e wave repeats itself only once, therefore Wavelength = L/2 = 16.5cm = 0.165 meters
D) The wave in figure b) is the 4th harmonic of wave in figure a). The frequnecy of wave in b) is given by
f4 = n*f1 where n is the hamonic number and f1 is the fundamental frequency.
we are given 4th harmonic f4 = 805 Hz
Therefore fundamental frequency of the voilin f1 = f4/ 4 = 805/4 = 201.25 Hz
E) The wave in figure c) is the second harmonic of the wave in figure b), i.e frequency of note in c)
f2 = 2*f1 = 2 * 201.25 = 402.5 Hz.
F) The general expression for the frequnecy of any note given by fn = n*f1 and
number of anitnodes is given by n(harmonic number)
G) The note in d is 8th harmonic, so fd = 8*f1 = 1610 Hz
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