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sergejj [24]
3 years ago
13

A small sphere of radius R is arranged to pulsate so that its radius varies in simple harmonic motion between a minimum of R−x a

nd a maximum of R+x with frequency f. This produces sound waves in the surrounding air of density rho and bulk modulus B.
a- Find the intensity of sound waves at the surface of the sphere. (The amplitude of oscillation of the sphere is the same as that of the air at the surface of the sphere.)


b-Find the total acoustic power radiated by the sphere


c-At a distance d≫R from the center of the sphere, find the amplitude of the sound wave.


d-At a distance d≫R from the center of the sphere, find the pressure amplitude of the sound wave.


e-At a distance d≫R from the center of the sphere, find the intensity of the sound wave.


Express your answer in terms of the variables R, x, f, and appropriate constants.
Physics
1 answer:
Colt1911 [192]3 years ago
5 0

Answer:

The intensity of sound wave at the surface of the sphere I =   \frac{ 2\pi^{2}R^{2} f^{2}\sqrt{\rho B}(\triangle R)^{2}}{ d^{2} }

Explanation:

B = Bulk modulus

Intensity, I = \frac{P_{max} ^{2} }{2\sqrt{\rho B} }

The amplitude of oscillation of the sphere is given by:

P_{max} = BkA\\k = \frac{2\pi }{\lambda} \\

A = \triangle R\\

Substitute v and A into Pmax

P_{max} = (2\pi f)\sqrt{\rho B} \triangle R\\ P_{max} ^{2} = 4\pi^{2} *f^{2} \rho B (\triangle R)^{2}

I = \frac{ 4\pi^{2} f^{2} \rho B (\triangle R)^{2}}{2\sqrt{\rho B} }

P_{total} = 4\pi R^{2} I

P_{total} =4\pi R^{2}  \frac{ 2\pi^{2} f^{2} \rho B (\triangle R)^{2}}{\sqrt{\rho B} }

The intensity of the sound wave at a distance  is given by:

I = \frac{P_{total} }{4\pi d^{2} }

I = 4\pi R^{2}  \frac{ 2\pi^{2} f^{2} \rho B (\triangle R)^{2}}{\sqrt{\rho B} } * \frac{1}{4\pi d^{2} } \\I =   \frac{ 2\pi^{2}R^{2} f^{2}\sqrt{\rho B}(\triangle R)^{2}}{ d^{2} }

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katrin2010 [14]

Answer:

There's a decrease in width of 2.18 × 10^(-6) m

Explanation:

We are given;

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Force;F = 60000 N.

Poisson’s ratio; υ =0.30

We are told width is 20 mm and thickness 40 mm.

Thus;

Area = 20 × 10^(-3) × 40 × 10^(-3)

Area = 8 × 10^(-4) m²

Now formula for shear modulus is;

E = σ/ε_z

Where σ is stress given by the formula Force(F)/Area(A)

While ε_z is longitudinal strain.

Thus;

E = (F/A)/ε_z

ε_z = (F/A)/E

ε_z = (60,000/(8 × 10^(-4)))/(207 × 10^(9))

ε_z = 3.62 × 10^(-4)

Now, formula for lateral strain is;

ε_x = - υ × ε_z

ε_x = -0.3 × 3.62 × 10^(-4)

ε_x = -1.09 × 10^(-4)

Now, change in width is given by;

Δw = w_o × ε_x

Where w_o is initial width = 20 × 10^(-3) m

So; Δw = 20 × 10^(-3) × -1.09 × 10^(-4)

Δw = -2.18 × 10^(-6) m

Negative means the width decreased.

So there's a decrease in width of 2.18 × 10^(-6) m

6 0
3 years ago
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Answer:

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Explanation:

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Now, to find the accurate number of significant figures when adding measurements, the basic rule for addition is to use the least number of decimal places when reporting the result.

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8 0
4 years ago
Two wires with the same resistance have the same diameter but different lengths. If wire 1 has length L 1 and wire 2 has length
SVEN [57.7K]

Answer with Explanation:

We are given that

Length of wire 1=L_1

Length of wire 2=L_2

Resistivity of copper wire=\rho_1=1.7\times 10^{-5}\Omega-m

Resistivity of aluminum wire=\rho_2=2.82\times 10^{-5}\Omega-m

Wire 1=Copper wire

Wire 2=Aluminum wire

Diameter of both wires are same and resistance of both wires are also same.

We know that

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\rho_1

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L_1=1.66L_2

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