Q.A source of sound of frequency $600\ \text{Hz}$ moves toward a stationary point at $10\ \text{m s}^{-1}$, while an observer at that point moves toward the source at $5\ \text{m s}^{-1}$. Taking the speed of sound in air as $340\ \text{m s}^{-1}$, find the frequency of sound heard by the observer.

Given: $f_0=600\ \text{Hz}$, $v_s=10\ \text{m s}^{-1}$ (source moving toward the point), $v_o=5\ \text{m s}^{-1}$ (observer moving toward the source), $v=340\ \text{m s}^{-1}$.

Both are approaching each other, so both corrections raise the apparent frequency; using the general formula:

$f' = f_0\,\frac{v+v_o}{v-v_s} = 600\times\frac{340+5}{340-10} = 600\times\frac{345}{330}$ …

Recognise that both source and observer approach each other, so take both $v_s$ and $v_o$ positive in the general formula $ …

• Applying only one of the two corrections (treating it as a source-only or observer-only problem) and ignoring the other body's motion. …