An ambulance is moving towards a stationary listener at a constant speed of 30 m s⁻¹ - NSC Physical Sciences - Question 6 - 2016 - Paper 1

Using the formula for wave speed:

$v = f \lambda$

we can rearrange to find the frequency:

$f = \frac{v}{\lambda}$

Substituting in the values:

$f_s = \frac{340}{0.28} = 1214.29 \text{ Hz}$

For the listener, we use the Doppler effect formula for sound when the source is moving towards a stationary observer:

$f' = f_s \frac{v + v_o}{v - v_s}$

Here, $f'$ is the frequency heard by the listener, $v$ is the speed of sound, $v_o$ is the listener's speed (0 m s⁻¹), and $v_s$ is the speed of the ambulance (30 m s⁻¹):

Substituting the values:

$f' = 1214.29 \frac{340 + 0}{340 - 30} = 1214.29 \frac{340}{310} \approx 1285.79 \text{ Hz}$

The frequency heard by the listener would DECREASE. This is because, as the speed of the ambulance decreases, the relative motion between the source and observer decreases, leading to a lower frequency being detected.

When observing the spectrum of a distant star, if the spectral lines are shifted towards the red end of the spectrum (redshift), it indicates that the star is moving away from the Earth. This effect occurs because the wavelengths of light emitted by the star stretch as it moves away, resulting in lower frequencies being detected compared to the frequencies when the star is stationary.