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

To calculate the frequency of the sound heard, we can use the Doppler effect formula:

$f' = f \left( \frac{v + v_{0}}{v - v_{s}} \right)$

where:

• $f' =$ frequency heard by the observer
• $f = 510 \, \text{Hz}$ (frequency of the source)
• $v = 340 \, \text{m/s}$ (speed of sound in air)
• $v_{0} = 0 \, \text{m/s}$ (speed of the observer, stationary person)
• $v_{s} = 12 \, \text{m/s}$ (speed of the source, car)

Plugging in the values:

$f' = 510 \left( \frac{340 + 0}{340 - 12} \right)$

Calculating:

$f' = 510 \left( \frac{340}{328} \right) \approx 511.5 \text{Hz}$

Thus, the frequency of the sound heard by the person as the car approaches is approximately 511.5 Hz.

Using the given Doppler effect relationship for blood cells:

$\Delta f = \frac{2f_{t} v_{r} \cos \theta}{c}$

Where:

• $\Delta f = 286 \, \text{Hz}$
• $f_{t} = 3.70 \times 10^{6} \, \text{Hz}$
• $c = 1540 \, \text{m/s}$
• $\theta = 60°$ (cos 60° = 0.5)

Rearranging the formula to find $v_{r}$ gives us:

$v_{r} = \frac{\Delta f \, c}{2 f_{t} \cos \theta}$

Substituting the known values:

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

Calculating:

$v_{r} = \frac{440840}{3.7 \times 10^{6}} \approx 0.119 \, \text{m/s}$

Therefore, the velocity of the red blood cells during this test is approximately 0.119 m/s.