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3 years ago

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An air-filled pipe is found to have successive harmonics at 980 Hz , 1260 Hz , and 1540 Hz . It is unknown whether harmonics bel

ow 980 Hz and above 1540 Hz exist in the pipe. What is the length of the pipe

1 answer:

zmey [24]3 years ago

3 0

Answer:

L = 0.7 m

Explanation:

This is a resonance exercise, in this case the air-filled pipe is open at both ends, therefore we have bellies at these points.

λ / 2 = L 1st harmonic

λ = L 2nd harmonic

λ = 2L / 3 3rd harmonic

λ = 2L / n n -th harmonic

the speed of sound is related to wavelength and frequencies

v =λ f

f = v /λ

we substitute

f = v n / 2L

the speed of sound in air is v = 343 m / s

suppose that the frequency of f = 980Hz occurs in harmonic n

f₁ = v n / 2L

f₂ = v (n + 1) / 2L

f₃ = v (n + 2) / 2L

we substitute the values

2 980/343 = n / L

2 1260/343 = (n + 1) / L

2 1540/343 = (n + 2) / L

we have three equations, let's use the first two

5.714 = n / L

7.347 = (n + 1) / L

we solve for L and match the expressions

n / 5,714 = (n + 1) / 7,347

7,347 n = 5,714 (n + 1)

n (7,347 -5,714) = 5,714

n = 5,714 / 1,633

n = 3.5

as the number n must be integers n = 4 we substitute in the first equation

L = n / 5,714

L = 0.7 m

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Answer:

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Explanation:

a) The vectorial equation of momentum is represented by the following expression:

$\vec p = m\cdot \vec v$ (1)

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$\vec p$ - Vector momentum, measured in kilogram-meters per second.

$m$ - Mass of the particle, measured in kilograms.

$\vec v$ - Vector velocity, measured in meters per second.

If we know that $m = 2.93\,kg$ and $\vec v = 2.98\,\hat{i}-3.98\,\hat{j}\,\,\,\left[\frac{m}{s} \right]$ , then the momentum is:

$\vec p = (2.93)\cdot (2.98\,\hat{i}-3.98\,\hat{j})\,\,\,\left[\frac{kg\cdot m}{s} \right]$

$\vec p = 8.731\,\hat{i}-11.661\,\hat{j}\,\,\,\left[\frac{kg\cdot m}{s} \right]$

The x and y components of the momentum are $8.731\,\frac{kg\cdot m}{s}$ and $-11.661\,\frac{kg\cdot m}{s}$ , respectively.

b) The magnitude and direction of momentum are represented by the following expressions:

$\|\vec p \| = \sqrt{p_{x}^{2}+p_{y}^{2}}$ (2)

$\theta = \tan^{-1}\left(\frac{p_{y}}{p_{x}} \right)$ (3)

Where:

$\|\vec p\|$ - Magnitude of momentum, measured in kilogram-meters per second.

$\theta$ - Direction of momentum, measured in sexagesimal degrees.

If we know that $p_{x} = 8.731\,\frac{kg\cdot m}{s}$ and $p_{y} = -11.661\,\frac{kg\cdot m}{s}$ , then the magnitude and direction of momentum are, respectively:

$\|\vec p\| = \sqrt{\left(8.731\,\frac{kg\cdot m}{s} \right)^{2}+\left(-11.661\,\frac{kg\cdot m}{s} \right)^{2}}$

$\|\vec p\| \approx 14.567\,\frac{kg\cdot m}{s}$

$\theta =\tan^{-1}\left(\frac{-11.661\,\frac{kg\cdot m}{s} }{8.731\,\frac{kg\cdot m}{s} } \right)$

$\theta \approx 306.823^{\circ}$

The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

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