What is the Doppler effect? Describe, with the aid of labelled diagrams, how the Doppler effect occurs - Leaving Cert Physics - Question b - 2022

To illustrate the Doppler effect, we can use diagrams showing concentric circles that represent wavefronts. When the wave source moves towards the observer, the waves become compressed, leading to shorter wavelengths and a higher observed frequency. Conversely, as the wave source moves away from the observer, the waves are stretched, resulting in longer wavelengths and a lower observed frequency. It is important to label the following in the diagrams:

• Direction of motion of the wave source
• Wavefronts approaching the observer
• Wavefronts moving away from the observer
• Differences in wavelength and frequency.

To calculate the frequency observed by Pierre, we first determine the velocity of the toy after it falls for 3 seconds due to gravity:

$v = u + at$ where:

• $u = 0$ ( (initial,velocity) )
• $a = 9.8 \, m/s^2$ ( (acceleration \ due \ to \ gravity) )
• $t = 3 \, s$ ( (time) )

Calculating the final velocity:

$v = 0 + (9.8 \, m/s^2)(3 \, s) = 29.4 \, m/s$

Next, we use the Doppler formula to find the observed frequency:

$f' = f \left( \frac{c}{c - u} \right)$ where:

• $f = 500 \, Hz$ ( (emitted frequency) )
• $c = 340 \, m/s$ ( (speed \ of \ sound) )
• $u = v = 29.4 \, m/s$ ( (relative \ speed) )

Substituting the values in, we find:

$f' = 500 \left( \frac{340}{340 - 29.4} \right)$

Calculating:

$f' = 500 \left( \frac{340}{310.6} \right) \approx 500 \times 1.093 \approx 546.5 \, Hz$

Thus, after 3 seconds, Pierre observes an approximate frequency of 546.5 Hz.