Wave Behaviour - Doppler effect

A police car siren emits a sound of frequency $1200\text{ Hz}$. The car is traveling at $25\text{ m s}^{-1}$ towards a stationary pedestrian. Calculate the frequency heard by the pedestrian. (Speed of sound $v = 340\text{ m s}^{-1}$)

Given: $f = 1200\text{ Hz}$ $v_s = 25\text{ m s}^{-1}$ $v = 340\text{ m s}^{-1}$

Using the moving source formula (moving towards): $f' = f \left( \frac{v}{v - v_s} \right)$ $f' = 1200 \times \left( \frac{340}{340 - 25} \right)$ $f' = 1200 \times \left( \frac{340}{315} \right)$ $f' \approx 1295.2\text{ Hz}$

Since the source is moving toward the observer, the denominator $(v - v_s)$ is smaller than $v$, which correctly results in an observed frequency higher than the source frequency.

Light from a distant galaxy is observed to have a wavelength of $658.2\text{ nm}$. The known laboratory wavelength for this specific spectral line is $656.3\text{ nm}$. Determine the velocity of the galaxy relative to Earth.

Given: $\lambda_{lab} = 656.3\text{ nm}$ $\lambda_{obs} = 658.2\text{ nm}$ $\Delta \lambda = 658.2 - 656.3 = 1.9\text{ nm}$

Using the light formula: $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$ $v = c \times \frac{\Delta \lambda}{\lambda}$ $v = (3.00 \times 10^8) \times \frac{1.9}{656.3}$ $v \approx 8.68 \times 10^5\text{ m s}^{-1}$

The observed wavelength is longer than the laboratory wavelength (red shift), indicating that the galaxy is moving away from Earth at approximately $868\text{ km s}^{-1}$.