Simple pendulum on Jupiter

SIMPLE PENDULUM ON JUPITER

INTRODUCTION

Jupiter is the fifth planet in our Solar System. Consider a simple pendulum on the surface of Jupiter ready to swing. A pendulum is a weight suspended from a pivot so that it can swing freely. It has a bob [mass] suspended from a frictionless pivot via a string. The central position of the bob i.e. when the pendulum is at rest is called its Mean position. When the bob is made to swing on the application of external force, it oscillates back and forth about this mean position. The maximum distance traversed by the bob from the mean position is known as amplitude. The amplitude is measured in radians which is a unit of angle. The time taken by the pendulum to complete one full oscillation is called the Time Period. For small amplitude less than 1 radian, the time period is independent of amplitude of the pendulum. We intend to determine the time period of such a pendulum on Jupiter.

A simple pendulum

ASSUMPTIONS

1. The string has no tension or compression

2. The pendulum is indestructible

3. Air resistance is negligible

4. Wind resistance is also negligible

5. Effect of Gravitational Time dilation is negligible

CALCULATION

The time period of a simple pendulum is given by,

T = 2π*√ (l/g) (Eqn. 1)

Where,

T – Time period [s]

l – Length of the pendulum [m]

g – Acceleration due to gravity on Jupiter [m/s²]

g = 24.5 m/s² (Eqn. 2) 

Let the length of pendulum be

l = 1 m (Eqn. 3)

Now substitute equations (2), (3) in equation (1)

T = 2π*√ (1/24.5)

T = 2π*√ (0.0408)

T = 2π*0.2020

T = 1.2693 s (Eqn. 4)

This is the time period of a simple pendulum on Jupiter. It means any pendulum performing small oscillations will always take approximately 1.3 second to complete one cycle anywhere on Jupiter provided the value of gravity is same everywhere and the effects of friction are neglected. Due to higher gravity of Jupiter, the time period of pendulum is less compared to Earth where it was 2 second.

CONCLUSION

We thus determined the time period of a simple pendulum on Jupiter.

Simple pendulum on Moon

SIMPLE PENDULUM ON MOON

INTRODUCTION

Moon is Earth’s natural satellite orbiting Earth at a distance of 400,000 Km. Consider a simple pendulum on the surface of Moon ready to swing. A pendulum is a weight suspended from a pivot so that it can swing freely. It has a bob [mass] suspended from a frictionless pivot via a string. The central position of the bob i.e. when the pendulum is at rest is called its Mean position. When the bob is made to swing on the application of external force, it oscillates back and forth about this mean position. The maximum distance traversed by the bob from the mean position is known as amplitude. The amplitude is measured in radians which is a unit of angle. The time taken by the pendulum to complete one full oscillation is called the Time Period. For small amplitude less than 1 radian, the time period is independent of amplitude of the pendulum. We intend to determine the time period of such a pendulum on Moon.

A simple pendulum

ASSUMPTIONS

1. The string has no tension or compression

2. The pendulum is indestructible

3. Effect of Gravitational Time dilation is negligible

CALCULATION

The time period of a simple pendulum is given by,

T = 2π*√ (l/g) (Eqn. 1)

Where,

T – Time period [s]

l – Length of the pendulum [m]

g – Acceleration due to gravity on Moon [m/s²]

g = 1.620 m/s² (Eqn. 2)

Let the length of pendulum be

l = 1 m (Eqn. 3)

Now substitute equations (2), (3) in equation (1)

T = 2π*√ (1/1.620)

T = 2π*√ (0.6172)

T = 2π*0.7856

T = 4.9365 s (Eqn. 4)

This is the time period of a simple pendulum on Moon. It means any pendulum performing small oscillations will always take approximately 5 second to complete one cycle anywhere on Moon provided the value of gravity is same everywhere and the effects of friction are neglected. Due to lower gravity of moon, the time period is larger compared to Earth where it was 2 second.

CONCLUSION

We thus determined the time period of a simple pendulum on Moon.

Simple pendulum on Earth

SIMPLE PENDULUM ON EARTH

INTRODUCTION

Our Earth is the third planet in our solar system. Consider a simple pendulum on the surface of Earth ready to swing. A pendulum is a weight suspended from a pivot so that it can swing freely. It has a bob [mass] suspended from a frictionless pivot via a string. The central position of the bob i.e. when the pendulum is at rest is called its Mean position. When the bob is made to swing on the application of external force, it oscillates back and forth about this mean position. The maximum distance traversed by the bob from the mean position is known as amplitude. The amplitude is measured in radians which is a unit of angle. The time taken by the pendulum to complete one full oscillation is called the Time Period. For small amplitude less than 1 radian, the time period is independent of amplitude of the pendulum. We intend to determine the time period of such a pendulum on Earth.

A simple pendulum

ASSUMPTIONS

1. The string has no tension or compression

2. The pendulum is indestructible

3. Air resistance is negligible

4. Effect of Gravitational Time dilation is negligible

CALCULATION

The time period of a simple pendulum is given by,

T = 2π*√ (l/g) (Eqn. 1)

Where,

T – Time period [s]

l – Length of the pendulum [m]

g – Acceleration due to gravity on Earth [m/s²]

g = 9.8 m/s² (Eqn. 2)

Let the length of pendulum be

l = 1 m (Eqn. 3)

Now substitute equations (2), (3) in equation (1)

T = 2π*√ (1/9.8)

T = 2π*√ (0.1020)

T = 2π*0.3194

T = 2.007 s (Eqn. 4)

This is the time period of a simple pendulum on Earth. It means any pendulum performing small oscillations will always take approximately 2 second to complete one cycle anywhere on Earth provided the value of gravity is same everywhere and the effects of friction are neglected.

CONCLUSION

We thus determined the time period of a simple pendulum on Earth.

Sonic Doppler shift in Military Jet Pt7

SONIC DOPPLER SHIFT IN MILITARY JET PT7

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 non-sonic jet is chasing a sonic military jet. A sonic military jet is a jet that moves at the speed of sound while a non-sonic jet moves at speeds less than that of sound. Consider two military jets, jet1 [source] moving at 343 m/s and jet2 [observer] moving at a speed of 300 m/s. We’ll determine the apparent frequency of source jet noise as registered by the observer in jet2 when jet1 is receding away from him.

ASSUMPTIONS

1    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    2. The effect of humidity on sound is negligible

3    3. The amplitude of sound is unity

4    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 [Jet2] (m/s)

V_s – Velocity of Source [Jet1] (m/s)

The Doppler shift equation for this case is,

f’ = f₀*{[V + V_o]/[V + V_s]} (Eq. 2)                                                                                                       

The ‘+’ sign in the numerator of equation (2) indicates that the observer is moving toward the source while the ‘+’ sign in the denominator indicates that the source is moving away from the observer.

The velocity of jet1 [Source] V_s = 343 m/s (Eq. 3)                                                                                

The velocity of jet2 [Observer] V_o = 343 m/s (Eq. 4)                                                                  

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

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

Substitute equations (3), (4), (5) and (6) in equation (2),

f’ = 1000*{[343 + 300]/[343 + 343]}

f’ = 937.31 Hz

This is the frequency of source noise as registered by observer [Jet2] in the jet when he is chasing the source [Jet1]. We can observe that the observer in Jet2 will hear lower frequency as compared to the original value due to the Doppler shift.

Difference in frequency = f’ – f₀

                                       = 937.31 – 1000

                                       = – 62.69 Hz

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

CONCLUSION

We thus determined the apparent frequency of source noise as registered by the observer due to Doppler shift and concluded that the observer will be able to hear lower value of frequency of jet noise when he follows the source jet1.

Sonic Doppler shift in Stealth Jet

SONIC DOPPLER SHIFT IN STEALTH 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 one sonic military jet is chasing a non-sonic jet. A sonic military jet is a jet that moves at the speed of sound while a non-sonic jet moves at speeds less than that of sound. Consider two military jets, jet1 [source] moving at 343 m/s and jet2 [observer] moving at a speed of 300 m/s. We’ll determine the apparent frequency of source jet noise as registered by the observer in jet2 when jet1 is chasing him.

A Stealth Jet

ASSUMPTIONS

1    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    2. The effect of humidity on sound is negligible

3    3. The amplitude of sound is unity

4    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 [Jet2] (m/s)

V_s – Velocity of Source [Jet1] (m/s)

The Doppler shift equation for this case is,

f’ = f₀*{[V – V_o]/[V – V_s]} (Eq. 2)                                                                                                   

The ‘–’ sign in the numerator of equation (2) indicates that the observer is moving away from the source while the ‘–’ sign in the denominator indicates that the source is moving toward the observer.

The velocity of jet1 [Source] V_s = 343 m/s (Eq. 3)                                                         

The velocity of jet2 [Observer] V_o = 343 m/s (Eq. 4)                                                          

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

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

Substitute equations (3), (4), (5) and (6) in equation (2),

f’ = 1000*{[343 – 300]/[343 – 343]}

f’ = ∞ Hz

This is the frequency of source noise as registered by observer [Jet2] in the jet when the source [Jet1] is chasing him. The ‘∞’ value of frequency indicates that time is zero. It means that the source is moving with the sound waves and hence the observer will never hear anything as frequency does not exist for him although he may be able to see the source.

Difference in frequency = f’ – f₀

                                       = ∞ – 1000

                                       = ∞ Hz

APPLICATION

Stealth aircrafts traveling at sonic speeds will be totally quiet with respect to the observer who is being followed due to the Doppler Effect. The fugitive or the enemy plane will never hear anything about an aircraft following them. This helps the stealth jet to make the move and quickly evade the enemy jet.

CONCLUSION

We thus determined the apparent frequency of source noise as registered by the observer due to Doppler shift and concluded that the observer will be able to hear nothing when he is being chased by the source.