Question 233: Waves

Question

When the observer moves towards a stationary source with velocity $\mathrm{V}_1$, the apparent frequency of emitted note is $F_1$. When observer moves away from the source with velocity $\mathrm{V}_1$, the apparent frequency is $\mathrm{F}_2$. If V is the velocity of sound in air and $F_1 / F_2=2$ then $V / V_1$ is equal to

Accepted Answer

To find the apparent frequency when the observer is moving, we use the formula:
$$ \mathrm{F}^{\prime} = \left[ \frac{\mathrm{V} \pm \mathrm{V}_0}{\mathrm{V} \mp \mathrm{V}_s} \right] \mathrm{F} $$
Here, $ \mathrm{V} $ is the speed of sound, $ \mathrm{V}_0 $ is the speed of the observer, and $ \mathrm{V}_s $ is the speed of the source.
In the problem, the source does not move, so $ V_s=0 $. The observer moves with speed $ V_1 $, so $ V_0 = V_1 $.
When the observer moves
towards
the source, the formula becomes:

$$ F_1 = \left[ \frac{V + V_1}{V} \right] F $$
When the observer moves
away
from the source, the formula becomes:

$$ F_2 = \left[ \frac{V - V_1}{V} \right] F $$
Now, divide $ F_1 $ by $ F_2 $:

$$ \frac{F_1}{F_2} = \frac{V + V_1}{V - V_1} $$
The problem gives $ \frac{F_1}{F_2} = 2 $. So:

$$ 2 = \frac{V + V_1}{V - V_1} $$
Now, solve for $ V $ in terms of $ V_1 $:
Multiply both sides by $ V - V_1 $:

$$ 2(V - V_1) = V + V_1 $$
Expand:

$$ 2V - 2V_1 = V + V_1 $$
Move all terms with $ V $ to one side and all with $ V_1 $ to the other side:

$$ 2V - V = 2V_1 + V_1 $$

$$ V = 3V_1 $$
Divide both sides by $ V_1 $:

$$ \frac{V}{V_1} = 3 $$

Answer

3

Answer

2

Answer

1

Answer

4

Waves

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