Question 394: Waves

Question

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 $$=330 \mathrm{~m} / \mathrm{s}$$.]

Accepted Answer

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

Answer

same in both the cases.

Answer

more in the second case than in the first case.

Answer

less in the second case than in the first case.

Answer

less than the actual frequency in both the cases.

Waves

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