A pendulum in a grandfather clock oscillates back and forth twice in one second what is it’s period
1 answer:
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Answer:
2(maximum), -2(minimum), -2(maximum).
Explanation:
V(t)= 2πcos πt--------------------------------------------------------------------------------(1).
Therefore, there is a need to integrate v(t) to get S(t).
S(t)= 2×sinπt + C ------------------------------------------------------------------------------(2).
Applying the condition given, we have s(0)= 0.
S(0)= 2sin ×π(0) + C.
Which means that; 0+C= 0. That is; C=0.
S(t)= 2 sin πt.
The mass moves to its highest positions at time,t=half(1/2=.5) and time,t=2.5.
Take note that; sin(π/2) = sin(5π/2) = 1 .
Also, the mass moves to its lowest position at time,t=(3/2); also, sin(3π/2) = -1.
Therefore, we have that 2 maximum; -2 minimum and -2 maximum.
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