Sonic doppler shift in Military Jet Pt2

SONIC DOPPLER SHIFT IN MILITARY JET PT2

INTRODUCTION

We know that Doppler Effect or Doppler shift occurs between a source and observer when they are in relative motion with respect to each other. In this case we’ll determine the Doppler shift that occurs when a sonic military jet is moving away from a stationary observer in a building. A sonic military jet is a jet that moves at the speed of sound. Consider a military jet moving at a speed of 343 m/s receding away from an observer who is inside a building. We’ll determine the apparent frequency of the jet’s noise as registered by the observer.

ASSUMPTIONS

1. The atmospheric air has standard temperature and pressure conditions

·         Temperature T = 298 K or 25°C or 77°F

·         Pressure = 1 bar = 10⁵ N/m²

2. The effect of humidity on sound is negligible

3. The amplitude of sound is unity

4. The air molecules do not move with respect to source and observer

CALCULATION

The equation for Doppler shift is given by,

f’ = f₀*{[V ± V_o]/[V ± V_s]} (Eq. 1)                                                                                                         

f₀ – Original frequency (Hz)

f’ – Apparent or observed frequency (Hz)

V – Velocity of Sound in air at standard temperature and pressure conditions (m/s) {V = 343 m/s}

V_o – Velocity of observer (m/s)

V_s – Velocity of Source [Jet] (m/s)

Since the observer is stationary,

V_o = 0 (Eq. 2)                                                                                                                                           

Substitute equation (2) in equation (1),

f’ = f₀*{[V]/[V + V_s]} (Eq. 3)

The ‘+’ sign in the denominator of equation (3) indicates that the source is receding away from the observer.

The velocity of jet V_s = 343 m/s (Eq. 4)                                                                                                   

Frequency of jet exhaust noise f₀ = 1000 Hz (Eq. 5)                                                                                

Speed of sound in air V = 343 m/s (Eq. 6)                                                                                                

Substitute equations (4), (5) and (6) in equation (3),

f’ = 1000*{343/[343 + 343]}

f’ = 500 Hz

This is the frequency of jet noise as registered by the stationary observer in a building when a military jet is receding away from him. We can observe that the frequency reduced to half its original value which means the observer will only hear half of the original frequency.

Difference in frequency = f’ – f₀

                                       = 500 – 1000

                                       = – 500 Hz

Negative sign indicates that apparent frequency is less than original but magnitude is always positive.

CONCLUSION

We thus determined the apparent frequency of jet noise as registered by the observer due to Doppler shift and concluded that the observer will be able to hear only half of the original frequency of jet noise when he watches the jet receding away from him.

Sonic doppler shift in Military Jet

SONIC DOPPLER SHIFT IN MILITARY JET

INTRODUCTION

We know that Doppler Effect or Doppler shift occurs between a source and observer when they are in relative motion with respect to each other. In this case we’ll determine the Doppler shift that occurs when a sonic military jet is moving toward a stationary observer in a building. A sonic military jet is a jet that moves at the speed of sound. Consider a military jet moving at a speed of 343 m/s approaching an observer who is inside a building. We’ll determine the apparent frequency of the jet’s noise as registered by the observer.

ASSUMPTIONS

1. The atmospheric air has standard temperature and pressure conditions

·         Temperature T = 298 K or 25°C or 77°F

·         Pressure = 1 bar = 10⁵ N/m²

2. The effect of humidity on sound is negligible

3. The amplitude of sound is unity

4. The air molecules do not move with respect to source and observer

CALCULATION

The equation for Doppler shift is given by,

f’ = f₀*{[V ± V_o]/[V ± V_s]} (Eq. 1)                                                                                                        

f₀ – Original frequency (Hz)

f’ – Apparent or observed frequency (Hz)

V – Velocity of Sound in air at standard temperature and pressure conditions (m/s) {V = 343 m/s}

V_o – Velocity of observer (m/s)

V_s – Velocity of Source [Jet] (m/s)

Since the observer is stationary,

V_o = 0 (Eq. 2)                                                                                                                                        

Substitute equation (2) in equation (1),

f’ = f₀*{[V]/[V – V_s]} (Eq. 3)                                                                                                                

The ‘–’ sign in the denominator of equation (3) indicates that the source is approaching the observer.

The velocity of jet V_s = 343 m/s (Eq. 4)                                                                                                 

Frequency of jet exhaust noise f₀ = 1000 Hz (Eq. 5)                                                                               

Speed of sound in air V = 343 m/s (Eq. 6)                                                                                               

Substitute equations (4), (5) and (6) in equation (3),

f’ = 1000*{343/[343 – 343]}

f’ = ∞ Hz

This is the frequency of sound as registered by the stationary observer in a building when a military jet approaches him. The ∞ value indicates that time is 0. It means that the object [jet] is traveling with sound waves, hence an observer will never hear any sound as there exists no sound to an observer. However the observer will hear sound when the jet moves away from him which we’ll discuss in the next post.

Difference in frequency = f’ – f₀

                                       = ∞ – 1000

                                       = ∞ Hz [does not exist]

CONCLUSION

We thus determined the apparent frequency as registered by the observer due to Doppler shift and concluded that the observer will not be able to hear anything even though he’ll be able to see a jet approaching him.

GRAVITATIONAL FIELD OF AN ELLIPTICAL SECTOR PLATE

GRAVITATIONAL FIELD OF AN ELLIPTICAL SECTOR PLATE

INTRODUCTION

Gravity is derived from the Latin word ‘gravitas’ meaning mass. The universal law of gravitation was coined by Sir Isaac Newton. According to the law, any two masses anywhere in the universe separated by a distance will attract each other. This force of attraction is proportional to the product of their masses and inversely proportional to square of the distance between them.

(Eq .1)

The distance between two masses can be finite or infinite, which is why gravitational force is referred to as long range force but is also the weakest force among all the other fundamental forces. All objects that have mass will attract other masses. This means that each mass has its own gravitational field just like Earth. So this implies that all objects will attract each other since they will have their own field. This is not evident on Earth since Earth’s gravitational field outweighs all other mass’s field and hence all objects no matter how massive are attracted toward the Earth. In this post we intend to determine the gravitational field of a two dimensional Elliptical sector plate and identify points where the plate’s own gravity is strong at some parts and weak at the other.  

ASSUMPTIONS

1. The thickness of the plate is negligible compared to its length and width.

2. The plate is not under the influence of an external gravitational field.

3. The plate is a homogeneous material.

4. All the mass is assumed to be concentrated at the center.

CALCULATION

Consider an Elliptical Sector plate of major axis ‘a’ [m] and minor axis ‘b’ [m], angle ‘θ’ [radian] and mass M [Kg]. We will first determine the center of gravity of this plate and then the magnitude of the gravitational field at points of interest.

Center of gravity

We already determined the center of gravity of sector of regular ellipse in the previous post. The center of gravity of a regular Elliptical Sector is,

Figure .1 Sector of an ellipse

Points of interest and their distances from center

Consider an arbitrary point X ≡ (x,y) anywhere on the rim of the sector. We intend to determine the magnitude of gravitational field at point X. It is important to note that point X must lie on rim of the sector. If it lies inside the sector, we have to subtract the offset distance from the rim to determine the accurate value of gravitational field.

The distance between point X and C.G. can be calculated by the distance formula,

Gravitational field

From Eq. 1, the force of attraction between a mass and its own surface is given by,

g = GM/R²

g – Acceleration due to gravity of the mass (m/s²)

G – Universal constant of gravitation

G = 6.67*10^-11 Nm²/Kg²

M – Mass of the object (Kg)

R – Distance between the centers of two masses (m)

G field at point X

The gravitational field beyond the surface is obtained by adding the additional distance,

d – Distance from the surface of the object to other object.

CONCLUSION

We determined the magnitude of gravitational field at one arbitrary point X. From the equation that we derived, we can determine the magnitude of g field at any point of interest as long as we know the coordinate and its distance from the center of gravity.

Center of Gravity of Sector of an Ellipse

CENTER OF GRAVITY OF SECTOR OF AN ELLIPSE

INTRODUCTION

The center of gravity COG of an object is the point of action of the gravitational force. It is also known as the balancing point since all objects [simple or complex] have their COG within the object’s sphere of influence and this ensures stability. There are two values to be noted which is the geometrical center of gravity and the actual center of gravity. The geometrical COG is the exact center of the object. It can be calculated by various available methods like summation, moment of inertia, etc. The geometrical COG is valid as long as there is uniform distribution of mass. In case of uneven mass distribution or use of composite or heterogeneous materials, the actual COG will no longer coincide with the geometrical COG. This is because mass is distributed unevenly and the COG will shift where there is more mass. In this article we intend to determine the COG of sector of an ellipse and we assume uniform mass distribution for simplicity.

ASSUMPTIONS

1. Mass of the object is evenly distributed

2. Earth has uniform gravitational field

CALCULATION

Consider a sector of ellipse inscribed inside an actual ellipse. The ellipse has major axis ‘2a’ and minor axis ‘2b’. Consider a small element [a sliver of the sector]. Let the angle of this element be ‘dθ’. Let ‘θ’ be the angle between this element and one of the side of the ellipse.

Figure .1 Sector of an Ellipse

To determine the center of gravity, we need to follow three steps:

1. Determine the area of the element

2. Determine the area of the final object in question [sector]

3. Determine the distance between COG of element and the origin

Once we have the above three values, we can determine the COG coordinates.

Step1: Area of object [sector]

The sector of an ellipse is just a subset of the entire ellipse, hence the area will also be a part of the entire ellipse area. The area will depend on the angle of the sector. The equation is as follows:

Step2: Area of the element

From the figure, the element can be approximated as a triangle with height ‘h’ and base ‘hdθ’ respectively. Now the approximate perimeter of ellipse is given by,

The base of the triangle element which is also the arc length of the ellipse is given by,

But arc length is also equal to hdθ, so on comparison from both equations we obtain that height,

The area of the element dA is as follows,

 

Step 3: COG of the element

Since the element is a triangle, the COG of a triangle is distributed in the ratio 2/3 and 1/3. The COG is (2/3)*h away from the origin. We then resolve the distance into two components which is x and y. We get cosine term for the x component while the sine term for y component.

Let (x_s ,y_s) be the COG coordinates of the element.

Now we can proceed to determine the COG of the object

The equation is,

Now substitute all values to find the center of gravity coordinates,

On performing numerical computation we get the coordinates as,

CONCLUSION

We thus determined the center of gravity of Sector of an ellipse.

Gravitational field of a Circular sector plate

GRAVITATIONAL FIELD OF A CIRCULAR SECTOR PLATE

INTRODUCTION

Gravity is derived from the Latin word ‘gravitas’ meaning mass. The universal law of gravitation was coined by Sir Isaac Newton. According to the law, any two masses anywhere in the universe separated by a distance will attract each other. This force of attraction is proportional to the product of their masses and inversely proportional to square of the distance between them.

(Eq. 1)

The distance between two masses can be finite or infinite, which is why gravitational force is referred to as long range force but is also the weakest force among all the other fundamental forces. All objects that have mass will attract other masses. This means that each mass has its own gravitational field just like Earth. So this implies that all objects will attract each other since they will have their own field. This is not evident on Earth since Earth’s gravitational field outweighs all other mass’s field and hence all objects no matter how massive are attracted toward the Earth. In this post we intend to determine the gravitational field of a two dimensional Circular sector plate and identify points where the plate’s own gravity is strong at some parts and weak at the other.   

ASSUMPTIONS

1. The thickness of the plate is negligible compared to its length and width.

2. The plate is not under the influence of an external gravitational field.

3. The plate is a homogeneous material.

4. All the mass is assumed to be concentrated at the center.

CALCULATION

Consider a Circular Sector plate of radius ‘r’ [m], angle ‘α’ [radian] and mass M [Kg]. We will first determine the center of gravity of this plate and then the magnitude of the gravitational field at points of interest.

Center of gravity

The center of gravity of a regular Sector is (x,y) ≡ (2rsinα/3α, 0)

Figure .1 Sector of a Circle

Points of interest and their distances from center

We will consider three points A, B and O as shown in figure 1.

Point A ≡ (r, 0)

The distance between Point A and C.G. can be calculated by the distance formula

Point B ≡ (r, rsinα)

The distance between Point B and C.G. can be calculated by the distance formula

Point O ≡ (0, 0)

The distance between Point O and C.G. can be calculated by the distance formula

Gravitational field

From Eq. 1, the force of attraction between a mass and its own surface is given by,

g – Acceleration due to gravity of the mass (m/s²)

G – Universal constant of gravitation

G = 6.67*10^-11 Nm²/Kg²

M – Mass of the object (Kg)

R – Distance between the centers of two masses (m)

G field at point A

The gravitational field beyond the surface is obtained by adding the additional distance,

d – Distance from the surface of the object to other object.

G field at point B

The gravitational field beyond the surface is obtained by adding the additional distance,

G field at point O

The gravitational field beyond the surface is obtained by adding the additional distance,

CONCLUSION

We thus determined the magnitude of gravitational field at various points on the Circular Sector plate. On performing numerical calculation to solve the coefficients of g field equations, point A turns out to be the point of strongest gravitational field. Point O follows point A in terms of field strength while point B has the weakest gravitational field due to its large distance from the center of gravity.