Question 316: Waves

Question

When the listener moves towards 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 ' $\mathrm{V}_1$ ', apparent frequency is ' $F_2$ '. If V is the velocity of sound in air and $\frac{F_1}{F_2}=2$ then $\frac{V}{V_1}$ is

Accepted Answer

Let's start by writing down the Doppler effect formulas for both cases:
When the observer moves toward the stationary source with speed $$V_1$$, the apparent frequency is:
$$F_1 = f 	imes \frac{V + V_1}{V},$$
where $$f$$ is the source frequency and $$V$$ is the speed of sound.
When the observer moves away from the source with speed $$V_1$$, the frequency becomes:
$$F_2 = f 	imes \frac{V - V_1}{V}.$$
The problem tells us that:
$$\frac{F_1}{F_2} = 2.$$
Substitute the expressions for $$F_1$$ and $$F_2$$:
$$\frac{F_1}{F_2} = \frac{f 	imes \frac{V + V_1}{V}}{f 	imes \frac{V - V_1}{V}} = \frac{V + V_1}{V - V_1} = 2.$$
Now, solve for $$V$$ in terms of $$V_1$$:
Write the equation:
$$\frac{V + V_1}{V - V_1} = 2.$$
Cross-multiply:
$$V + V_1 = 2(V - V_1).$$
Expand the right-hand side:
$$V + V_1 = 2V - 2V_1.$$
Rearrange the equation to isolate terms with $$V$$ and $$V_1$$:
$$V + V_1 - 2V + 2V_1 = 0,$$
$$-V + 3V_1 = 0.$$
Solve for $$V$$:
$$V = 3V_1.$$
Thus, the ratio is:
$$\frac{V}{V_1} = 3.$$
So, the correct answer is Option B.

Answer

,2

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3

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4

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5

Waves

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