5. (a) A person is standing at the side of a road - Scottish Highers Physics - Question 5 - 2019

To calculate the frequency heard by the observer, we can use the Doppler effect formula:

$f_{0} = f \frac{v + v_{s}}{v}$

Substituting the values:

• $f = 510 \, \text{Hz}$ (the frequency of the sound emitted by the car)
• $v = 340 \, \text{m/s}$ (the speed of sound in air)
• $v_{s} = 12 \, \text{m/s}$ (the speed of the car approaching the observer)

Thus,

$f_{0} = 510 \frac{340 + 12}{340}$

Calculating:

$f_{0} = 510 \frac{352}{340} \approx 530 \, \text{Hz}$

Using the relationship for the change in frequency:

$\Delta f = \frac{2 f_{t} v_{k} \cos \Theta}{v}$

Substituting the known values:

• $\Delta f = 286 \, \text{Hz}$
• $f_{t} = 3.70 \times 10^{6} \, \text{Hz}$ (as 1 MHz = $10^6$ Hz)
• $v = 1540 \, \text{m/s}$
• $\Theta = 60^{\circ}$

Now we apply these in the equation:

$286 = \frac{2 \times 3.70 \times 10^{6} \times v_{k} \cos(60^{\circ})}{1540}$

Since $\cos(60^{\circ}) = 0.5$:

$286 = \frac{2 \times 3.70 \times 10^{6} \times v_{k} \times 0.5}{1540}$

Rearranging for $v_{k}$:

$v_{k} = \frac{286 \times 1540}{2 \times 3.70 \times 10^{6} \times 0.5}$

Calculating:

$v_{k} \approx 119 \, \text{ms}^{-1}$