In what direction would the particles in this wave move, relative to the direction of wave travel?
2 answers:
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Answer:
the required minimum length of the attenuator is 3.71 cm
Explanation:
Given the data in the question;
we know that;
= c / 2a
where f is frequency, c is the speed of light in air and a is the plate separation distance.
we know that speed of light c = 3 × 10⁸ m/s = 3 × 10¹⁰ cm/s
plate separation distance a = 7.11 mm = 0.0711 cm
so we substitute
= 3 × 10¹⁰ / 2( 0.0711 )
= 3 × 10¹⁰ cm/s / 0.1422 cm
= 21.1 GHz which is larger than 15 GHz { TEM mode is only propagated along the wavelength }
Now, we determine the minimum wavelength required
Each non propagating mode is attenuated by at least 100 dB at 15 GHz
so
Attenuation constant TE₁ and TM₁ expression is;
∝₁ = 2πf/c × √( ( / f)² - 1 )
so we substitute
∝₁ = ((2π × 15)/3 × 10⁸ m/s) × √( (21.1 / 15)² - 1 )
∝₁ = 3.1079 × 10⁻⁷
∝₁ = 310.79 np/m
Now, To find the minimum wavelength, lets consider the design constraint;
20log₁₀ = -100dB
we substitute
20log₁₀ = -100dB
= 3.71 cm
Therefore, the required minimum length of the attenuator is 3.71 cm
Answer:
The maximum speed of sonic at the bottom of the hill is equal to 19.85m/s and the spring constant of the spring is equal to (497.4xmass of sonic) N/m
Energy approach has been used to sole the problem.
The points of interest for the analysis of the problem are point 1 the top of the hill and point 2 the bottom of the hill just before hitting the spring
The maximum velocity of sonic is independent of the his mass or the geometry. It is only depends on the vertical distance involved
Explanation:
The step by step solution to the problem can be found in the attachment below. The principle of energy conservation has been applied to solve the problem. This means that if energy disappears in one form it will appear in another.
As in this problem, the potential and kinetic energy at the top of the hill were converted to only kinetic energy at the bottom of the hill. This kinetic energy too got converted into elastic potential energy .
x = compression of the spring = 0.89
