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1) A piano tuner is testing a wire currently under 2000 N tension. The wire is 1.5 m long and has a mass of 20 g. If the tuner's computer can detect a maximum frequency of 10,000 Hz, what is the highest harmonic number the computer can detect (1 point)
A) 77
B) 100
C) 25
D) 63
2) A wave is traveling with a speed of 400 m/s and it has a wavelength of 28 m. What will the frequency of the wave be if the wavelength is doubled but the speed remains the same? (1 point)
A) 9.09 Hz
B) 14.3 Hz
C) 21.7 Hz
D) 7.14 Hz
3) If a standing wave is vibrating in it's third harmonic, how many nodes does it have? (1 point)
A) 4
B) 3
C) 2
D) 1
4) An organ pipe, which is a pipe closed at one end, is 2.46 m long and the speed of sound in air in that pipe is 345 m/s. What is the frequency of the fundamental harmonic in this pipe? (1 point)
A) 44.9 Hz
B) 20.6 Hz
C) 35.1 Hz
D) 13.2 Hz
5) Using the same organ pipe from question 11, how many possible harmonics are there in the normal human hearing range of 20 Hz to 20,000 Hz? (1 point)
A) 570
B) 985
C) 101
D) 47
6) A standing wave is set up on a violin string to produce a note of A sharp Contra (58 Hz). The violin string is 32 cm long. If the sound is produced in the fundamental frequency, what is the speed of the wave moving across the string? (1 point)
A) 52.78 m/s
B) 18.83 m/s
C) 20.01 m/s
D) 37.12 m/s

Respuesta :

1.

Answer:

N = 77

Explanation:

1) As we know that Tension in the wire is given as

T = 2000 N

now mass per unit length of the wire is given as

[tex]\frac{m}{L} = \frac{20 \times 10^{-3} kg}{1.5 m}[/tex]

[tex]\frac{m}{L} = 0.013 kg/m[/tex]

Now fundamental frequency is given as

[tex]f_o = \frac{1}{2L}\sqrt{\frac{T}{m/L}}[/tex]

[tex]f_o = \frac{1}{2\times 1.5}\sqrt{\frac{2000}{0.013}}[/tex]

[tex]f_o = 129 Hz[/tex]

now highest frequency is given as 10000 Hz

so total harmonics is given as

[tex]N = \frac{10000}{129}[/tex]

[tex]N = 77 [/tex]

2.

Answer:

D) 7.14 Hz

Explanation:

As we know that

speed = (frequency)(wavelength)

given that

speed = 400 m/s

wavelength = 28 m

so frequency is given as

[tex]f = \frac{400}{28}[/tex]

[tex]f = 14.3 Hz[/tex]

now if wavelength is doubled then we will have

[tex]wavelength = 56 m[/tex]

new frequency will be

[tex]f = \frac{400}{56}[/tex]

[tex]f = 7.14 hz[/tex]

3.

Answer:

Number of nodes = 4

A) 4

Explanation:

Number of nodes = number of harmonics + 1

so here we know that

number of harmonics = 3

Number of nodes = 3 + 1

Number of nodes = 4

4.

Answer:

C) 35.1 Hz

Explanation:

fundamental frequency in an organ pipe closed at one end is given as

[tex]f = \frac{v}{4L}[/tex]

here we know that

v = 345 m/s

L = 2.46 m

now we have

[tex]f = \frac{345}{4(2.46)}[/tex]

[tex]f = 35.1 Hz[/tex]

5.

Answer:

A) 570

Explanation:

Number of harmonics is given as

[tex]N = \frac{maximum :\frequency}{fundamental\: frequency}[/tex]

[tex]N = \frac{20,000}{35.1}[/tex]

[tex]N = 570[/tex]

6.

Answer:

[tex]v = 37.12 m/s[/tex]

Explanation:

For fundamental frequency we know that

[tex]f_o = \frac{v}{2L}[/tex]

here we know that

[tex]f_o = 58 Hz[/tex]

[tex]L = 32 cm[/tex]

now we have

[tex]58 = \frac{v}{2(0.32)}[/tex]

[tex]v = 37.12 m/s[/tex]

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