The alarm of a vehicle parked next to a straight horizontal road goes off, emitting sound with a wavelength of 0,34 m - NSC Physical Sciences - Question 6 - 2018 - Paper 1

The patrol car is driving AWAY FROM the parked vehicle. This is concluded because the driver of the patrol car hears a sound frequency that is 50 Hz lower than the frequency emitted by the alarm, which indicates that the source of the sound (the alarm) is moving away from the observer (the patrol car).

We can use the formula for the Doppler effect:

$f_o = f_s \frac{v}{v + v_s}$

Where:

• $f_o$ is the observed frequency (frequency heard by the observer, 1000 Hz)
• $f_s$ is the source frequency (to be calculated)
• $v$ is the speed of sound in air (340 m/s)
• $v_s$ is the speed of the source (the patrol car). Here, since the observed frequency is 50 Hz lower than the emitted frequency, $f_o = f_s - 50 Hz$.

Setting up the equation, we have:

$1000 = f_s \frac{340}{340 + v_s}$

From the above, we can infer that:

$f_s = 1000 + 50 = 1050 \text{ Hz}$

The speed of the patrol car can be derived from the frequency calculation. Assuming the speed of the patrol car is 17 m/s:

Distance = Speed × Time,

so,

$x = v \times t$ $x = 17 \times 10 = 170 \text{ metres}$

Thus, the calculated distance x is 170 metres.