# A driver in a car, approaching a vertical wall notices that the frequency

Question:

A driver in a car, approaching a vertical wall notices that the frequency of his car horn, has changed from $440 \mathrm{~Hz}$ to $480 \mathrm{~Hz}$, when it gets reflected from the wall. If the speed of sound in air is $345 \mathrm{~m} / \mathrm{s}$, then the speed of the car is

1. $36 \mathrm{~km} / \mathrm{hr}$

2. $24 \mathrm{~km} / \mathrm{hr}$

3. $18 \mathrm{~km} / \mathrm{hr}$

4. $54 \mathrm{~km} / \mathrm{hr}$

Correct Option: 4,

Solution:

$\mathrm{f}_{1}=$ frequency heard by wall $=\mathrm{f}_{\mathrm{s}}=\left(\frac{\mathrm{v}_{\mathrm{s}}}{\mathrm{v}_{\mathrm{s}}-\mathrm{v}_{\mathrm{c}}}\right)$

$\mathrm{f}_{2}=$ frequency heard by driver after reflection from wall

$\mathrm{f}_{2}=\left(\frac{\mathrm{v}_{\mathrm{s}}+\mathrm{v}_{\mathrm{c}}}{\mathrm{v}_{\mathrm{s}}}\right) \mathrm{f}_{1}=\left(\frac{\mathrm{v}_{\mathrm{s}}+\mathrm{v}_{\mathrm{c}}}{\mathrm{v}_{\mathrm{s}}-\mathrm{v}_{\mathrm{c}}}\right) \mathrm{f}_{0}$

$\frac{\mathrm{f}_{2}}{\mathrm{f}_{0}}=\frac{\mathrm{v}_{\mathrm{s}}-\mathrm{v}_{\mathrm{c}}}{\mathrm{v}_{\mathrm{s}}+\mathrm{v}_{\mathrm{c}}}$

$\frac{48}{44}=\frac{v_{s}-v_{c}}{v_{s}+v_{c}}$

$12\left(\mathrm{v}_{\mathrm{s}}+\mathrm{v}_{\mathrm{c}}\right)=11\left(\mathrm{v}_{\mathrm{s}}-\mathrm{v}_{\mathrm{c}}\right)$

$23 \mathrm{v}_{\mathrm{c}}=\mathrm{v}_{\mathrm{s}}$

$\mathrm{v}_{\mathrm{c}}=\frac{\mathrm{v}_{\mathrm{s}}}{23}=\frac{345}{23}=15 \mathrm{~m} / \mathrm{s}$

$=\frac{15 \times 18}{5}=54 \mathrm{~km} / \mathrm{hr}$

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