3· The \"chromatic\" musical scale uses twelve steps (increasing frequencies) to

Question

3· The "chromatic" musical scale uses twelve steps (increasing frequencies) to transition smoothly through a full octave (a doubling of frequency). In other words, if you play any key on a standard piano, and then you play the twelve subsequent keys (every one -black and white-in order), moving to the right up the keyboard, the last note you play will be an octave higher than the note you first chose 220.000 233.082 246.942 261.626 277.183 293.665 311.127 329.628 349.228 369.994 391.995 415.305 440.000 A# An example is listed here (at right), where you choose A, as your starting note; that's the “A" key nearest but below (left of) middle C", on the piano. Note how each successive frequency is an increase by the same factor (the same multiplier): 21.0594631 C# D# a. For this problem, your first task is to form an Amajor7 chord" with these four notes:E To do this, complete the designs of four simple devices that are partly described (in no particular order) in the table below-one note per device (and you may design them assuming that the local speed of sound is 343 m/s). Each device should be a simple, straight instrument capable of creating standing waves. (You may specify theA length of each instrument to within 1 mm, so the notes will be close but approximate.) G# In other words, fill in the blanks to complete the table below. Show all work in the extra space, as needed

Explanation / Answer

3. for the given table

row 1

nodes = 3

antinodes = 2

hence the device is closed on both sides

now,

lambda = L for 3 nodes

hence

f = v/lambda = 343/L

type and speed : logntidinal, 343 m/s

Device : tube of air, closed on both sides

length : 1.237 m

frequency, f = 277.283751 Hz

Note : C#4

row 2

dfevice is open at both ends

L = 0.413 m

nodes = 1

hence

antinodes = 2

lambda = 2L

f = v/2L

hence

type and speed : logntidinal, 343 m/s

Device : tube of air, open on both sides

length : 0.413 m

frequency, f = 415.2542 Hz

Note : G#4

row 3

L = 0.413 m

nodes = 2

antinodes = 2

hence dfevice is open at one

lambda = 4L/3

f = 3v/4L

also, final note has to be A3, so f = 220 Hz

hence

L = 1.16931 m

hence

type and speed : logntidinal, 343 m/s

Device : tube of air, open on one sides

length : 1.169318 m

frequency, f = 220 Hz

Note : A4

row 4

speed of sound in medium, v = 150 m/s

hence its not air, so it is a string fixed at both ends

hence its transcerse wave travelling in the material

antinodes = 5

hence

nodes = 6

so, lambda = 5L/2

f = 2v/5L

but note is E4

hence

f = 329.628 Hz

hence

L = 0.1820233718 m

type and speed : transverse, 150 m/s

Device : string fixed on both sides

length : 0.1820233 m

frequency, f = 329.628 Hz

Note : E4

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.