Answer:
Step-by-step explanation:
Una ambulancia se desplaza hacia el norte a 23,5 m/s. Su sirena emite un sonido de 500Hz de frecuencia. Calcular la frecuencia a
parente que percibe un observador que viaja en su auto, también hacia el norte , a 28,6, cuando:
1 answer:
Answer:
(I attach your complete question in the picture below)
The apparent frequency would be
f' = 493.26 Hz
Step-by-step explanation:
This a Doppler's effect problem. This effects produces a slight change in the frequency of a traveling wave, due to the effect of the velocity of the receiver (observer).
The formula used to describe the change, as both vehicles are moving away of each other:
f' = [ (Vwave + Vobserver) / (Vwave + Vsource ) ]*f
Where
Vwave = 340 m/s ( speed of sound)
Vobserver = 18.6 m/s (observer = car)
Vsource = 23.5 m/s (speed of the ambulance = source)
f = 500 Hz (frequency of the sound)
We just need to substitute in the equation
f' = [(340+ 18.6) / (340+ 23.5)] * 500 Hz
f' = [0.9865] * 500 Hz = 493.26 Hz
f' = 493.26 Hz
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Answer:
Step-by-step explanation:
Given the mass is m =16kg, and 1N force will stretch the spring 1 m.
That is, F =1N,Z =1m. Now find the spring constant k:
F = kL = 1 = k(1) = k= 1N/m.
The damping force is 8times the instantaneous velocity, this means β = 8,
and the external force is f(t) = 0
Initially the object compressed 0.6m above equilibrium position,
with the downward velocity is 2m/s.
The differential equation for a spring mass system with
damping force and extemal force is: mx" + βxt + kx = f(t).
so, 16x"+ 8x' + x= 0, x(0} = -0.6, x'(0)= 2m/s.
Now solve the DE:
The auxilary equation for the homogeneous equation is 16x"+8x'+x=0
solving we get, 16r² + 8r + 1 = 0 => (4r + 1)² = 0 => r = - 1/4.
Then the general solution for the homogenous system is: .
Use the initial conditions x (0) = -0.6, x'(0) = 2m/s:
.
Hence, .