Consider the Doppler effect in two cases. In the first case, an observer moves towards a stationary source…
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 $=330 \mathrm{~m} / \mathrm{s}$.]
same in both the cases.
more in the second case than in the first case.
less in the second case than in the first case.
less than the actual frequency in both the cases.
Solution
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.