Physics, 06.10.2019 08:30 diamondalize21p84czi

Acar is stopped at a traffic light. when the light turns green, the car starts from rest with an acceleration of 2.5 m/s2 . also, as the light turns green, a truck traveling with a constant velocity of 10 m/s passes the car. the truck’s horn is stuck, emitting sound at a frequency of 440hz. what frequency of sound is heard by the driver of the car, (a) 5 seconds before the light turned green? (b) 5 seconds after the light turned green? (c) 10 seconds after the light turned green?

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Answer from: jamalnellum56

Doppler effect is used to find the change in wave frequency when either the observer or source or both are moving

(a) 5 seconds before the light turned green, the frequency of sound heard by the observer is 453. $\overline {213}$ Hz

(b) 5 seconds after the light turned green the frequency of the sound heard by the observer is approximately 443.12 Hz

The reason the above values are correct is given as follows;

The given parameters are;

Initial velocity of the car = 0 (Starts from rest)

Acceleration of the car, a = 2.5 m/s²

The constant velocity of the truck that passes the car, v_s = 10 m/s

Frequency of the sound emitted by the truck, f = 440 Hz

(a) The frequency of the sound heard by the driver of the car 5 seconds before the light turns green is given as follows;

The Doppler shift formula is given as follows;

$f' = \dfrac{v }{v - v_s} \times f$

Where;

v = The speed of sound in air ≈ 343 m/s

Therefore;

$f' = \dfrac{343}{343 - 10} \times 440 \approx 453.\overline{213}$

The frequency of sound heard by the observer, f' = 453. $\overline {213}$ Hz

(b) The speed of the car t = 5 seconds after the light turns green is given as follows;

v = u + a·t

∴ v = 0 + 2.5 × 5 = 12.5

The speed of the car, 5 seconds after the light turns green, v_0 = 12.5 m/s

The position of the car at 5 seconds after the light turns green is given as follows;

$s = u\cdot t + \dfrac{1}{2} \cdot a \cdot t^2$

u = 0

$\therefore s = \dfrac{1}{2} \times 2.5 \times 5^2 = 31.25$

The position of the truck after 5 seconds = 5 s × 10 m/s = 50 m

Therefore, the car is moving towards the truck, which is moving away from the car

The correct formula to use is therefore;

$f' = \dfrac{v + v_0}{v + v_s} \times f$

$f' = \dfrac{343 + 12.5}{343 + 10} \times 440 \approx 443.12$

The frequency of the sound heard by the driver, f' ≈ 443.12 Hz

v = 0 + 2.5 × 10 = 25

The velocity of the car = 25 m/s

The position of the car, s₁₀, is given as follows;

$s_{10} = \dfrac{1}{2} \times 2.5 \times 10^2 = 125$

The position of the car, s₁₀ = 125 m

The position of the truck, d₁₀ = 10 s × 10 m/s = 100 m

Therefore, the car is ahead of the truck, and the equation to be used is therefore;

$f' = \dfrac{v - v_0}{v - v_s} \times f$

Which gives;

$f' = \dfrac{343 - 25}{343 - 10} \times 440 \approx 420.\overline{180}$

The frequency of the sound heard by the driver, f' = 420. $\overline{180}$ Hz

Learn more about the Doppler effect here:

link

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