A siren on the top of the police car moving with a speed of 40 m/s emits sound at frequency of 820Hz. Ahead of the car, there is a truck moving in the same direction as the car but with a speed of 30 m/s. Assume the speed of sound in air to be 340m/s.
a) What frequency does the driver of the truck hears as the police car approaches it?
b) What frequency does the driver of the truck hears after the police car has passed it and is receding from it?
a. Fd = ((Vs+Vd)/(Vs-Vc)) * Fc.
Fd = ((340+30)/(340-40)) * 820 =
b. Fd = ((340-30)/(340+40)) * 820 =
So the frequency heard by the driver will be greater than 820 Hz as the police car
approaches but less than 820 Hz as the car passes the truck.
To answer these questions, we need to consider the concept of the Doppler effect. The Doppler effect is the change in frequency or wavelength of a wave as observed by an observer moving relative to the source of the wave. It occurs for all types of waves, including sound waves.
Let's break down each question separately:
a) What frequency does the driver of the truck hear as the police car approaches it?
When the police car approaches the truck, the sound waves emitted by the siren will be compressed, resulting in an increased frequency observed by the truck driver.
To calculate the observed frequency, we use the formula for the Doppler effect:
f' = (v + vo) / (v + vs) * f
Where:
f' is the observed frequency
f is the frequency of the source (820 Hz in this case)
v is the speed of sound in air (340 m/s)
vo is the speed of the observer (driver of the truck, which is 0 m/s)
vs is the speed of the source (police car, which is 40 m/s)
Plugging in the values:
f' = (340 + 0) / (340 + 40) * 820
f' = 340 / 380 * 820
f' = 0.8947 * 820
f' ≈ 732.6 Hz
Therefore, the driver of the truck hears a frequency of approximately 732.6 Hz as the police car approaches.
b) What frequency does the driver of the truck hear after the police car has passed it and is receding from it?
When the police car has passed the truck and is receding from it, the sound waves emitted by the siren will be stretched, resulting in a decreased frequency observed by the truck driver.
We can use the same formula as before, but this time the speed of the source (vs) will be negative to indicate the opposite direction:
f' = (v + vo) / (v - vs) * f
Where:
f' is the observed frequency
f is the frequency of the source (820 Hz)
v is the speed of sound in air (340 m/s)
vo is the speed of the observer (0 m/s)
vs is the speed of the source (police car, which is -40 m/s)
Plugging in the values:
f' = (340 + 0) / (340 - (-40)) * 820
f' = 340 / 380 * 820
f' = 0.8947 * 820
f' ≈ 732.6 Hz
Therefore, the driver of the truck also hears a frequency of approximately 732.6 Hz after the police car has passed and is receding.
In conclusion, both before and after the police car passes the truck, the driver of the truck hears a frequency of approximately 732.6 Hz.
this is a straight forward doppler shift
just plug the information into your formulas