Question

When an observer moves towards a stationary source with velocity 'V₁', the apparent frequency of emitted note is 'F₁'. When observer moves away from stationary source with velocity 'V₁' the appearent frequency is 'F2'. If 'v' is velocity of sound in air and \[\frac {F_1}{F_2}\] = 2, then \[\frac {V}{V_1}\] is equal to ______.

Options

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Answer 1

When an observer moves towards a stationary source with velocity 'V₁', the apparent frequency of emitted note is 'F₁'. When observer moves away from stationary source with velocity 'V₁' the appearent frequency is 'F2'. If 'v' is velocity of sound in air and \[\frac {F_1}{F_2}\] = 2, then \[\frac {V}{V_1}\] is equal to 3.

Explanation:

Apparent frequency is given by,

F' = \[\left[{\frac{\mathrm{V}\pm\mathrm{V}_{0}}{\mathrm{V}\mp\mathrm{V}_{\mathrm{s}}}}\right]\mathrm{F}\]

Since source is stationary,

\[\therefore\] V_S = 0; V₀ = V₁

\[\therefore\] \[\mathrm{F}_1=\left[\frac{\mathrm{V}+\mathrm{V}_1}{\mathrm{V}}\right]\mathrm{F}\], \[\mathrm{F}_{2}=\left[\frac{\mathrm{V}-\mathrm{V}_{1}}{\mathrm{V}}\right]\mathrm{F}\]

\[\therefore\] \[\frac{\mathrm{F}_1}{\mathrm{F}_2}=\frac{\mathrm{V}+\mathrm{V}_1}{\mathrm{V}-\mathrm{V}_1}\]

\[\therefore\] 2 = \[\frac{\mathrm{V}+\mathrm{V}_{1}}{\mathrm{V}-\mathrm{V}_{1}}\]

\[\therefore\] 2V - 2V₁ = V + V1

\[\therefore\] V = 3V₁

\[\therefore\] \[\frac {V}{V_1}\] = 3