Consider the Doppler effect in two cases. In the first case, an observer moves towards a stationary source of sound with a speed of \(50 \mathrm{~m} / \mathrm{s}\). In the second case, the observer is at rest and the source moves towards the observer with the same speed of \(50 \mathrm{~m} / \mathrm{s}\). Then the frequency heard by the observer will be [velocity of sound in air \(=330 \mathrm{~m} / \mathrm{s}\).]

For observer moving towards a stationary source,
\(\mathrm{n}_1=\mathrm{n}_0\left[\frac{\mathrm{v}+\mathrm{v}_{\mathrm{L}}}{\mathrm{v}}\right]\)
For source moving towards a stationary observer,
\(\mathrm{n}_2=\mathrm{n}_0\left[\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right]\)
Substituting the values for \(v, v_L\) and \(v_S\) in the equations above
\(\begin{aligned}
& \mathrm{n}_1=\mathrm{n}_0\left[\frac{330+50}{330}\right]=1.15 \mathrm{n}_0 \\
& \mathrm{n}_2=\mathrm{n}_0\left[\frac{330}{330-50}\right]=1.17 \mathrm{n}_0 \\
& \Rightarrow \mathrm{n}_2>\mathrm{n}_1
\end{aligned}\)
\(\therefore \quad\) The frequency heard will be more in the second case than in the first case.