Answer
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Hint:To answer this question, we first need to understand what is pitch. A frequency's sensation is generally referred to as a sound's pitch. A high frequency sound wave corresponds to a high pitch sound, whereas a low frequency sound wave corresponds to a low pitch sound.
Complete step by step answer:
As we discussed above the frequency of a sound is directly proportional to its pitch, and the frequency of a sound is directly proportional to its velocity. Relation between temperature and velocity of sound in air is,
$\dfrac{{{V_T}}}{{{V_0}}} = \sqrt {\dfrac{T}{{{T_0}}}} $
(Here ${V_T}$ is the velocity of air at temp $T$(in kelvin) and ${V_0}$ is the velocity of air at temperature ${T_0}$(in kelvin))
As given in the question ${V_0}$= 332 m/s and $T$= ${20^0}C$ and ${T_0}$= ${0^0}C$.
Converting temperatures in kelvin
$T$= 20 + 273 = 293K
$\Rightarrow {T_0}$= 0 + 273 =273K
Substituting the given values
$\dfrac{{{V_T}}}{{332}} = \sqrt {\dfrac{{293}}{{273}}} $
$\Rightarrow \dfrac{{{V_T}}}{{332}} = 1.035$
Therefore ${V_T} = 343.62\,m/s$.
Doppler effect: The Doppler Effect is the shift in wave frequency that occurs as a wave source moves relative to its observer. When a sound source approaches you, for example, the frequency of the sound waves changes, resulting in a higher pitch. Applying Doppler effect when whistle is approaching observer
$\dfrac{{{f_a}}}{{{f_0}}} = \dfrac{{{V_T}}}{{{V_T} - {V_W}}}$
When whistle is distancing from observer,
$\dfrac{{{f_d}}}{{{f_0}}} = \dfrac{{{V_T}}}{{{V_T} + {V_W}}}$
Dividing these equations, we get,
$\dfrac{{{f_d}}}{{{f_a}}} = \dfrac{{{V_T} - {V_W}}}{{{V_T} + {V_W}}}$
As given in the question $\dfrac{{{f_d}}}{{{f_a}}} = \dfrac{4}{5}$
$\dfrac{4}{5} = \dfrac{{{V_T} - {V_W}}}{{{V_T} + {V_W}}}$ and by equating further we get
$\Rightarrow {V_W} = \dfrac{{{V_T}}}{9}$
$\therefore {V_W} = 38.2\,m/s$.........(By substituting value of ${V_T} = 343.62\,m/s$)
Hence, the correct answer is option C.
Note:Molecules with more energy can vibrate faster at higher temperatures. Sound waves can fly faster because molecules vibrate faster. In room temperature air, sound travels at 346 metres per second.
Complete step by step answer:
As we discussed above the frequency of a sound is directly proportional to its pitch, and the frequency of a sound is directly proportional to its velocity. Relation between temperature and velocity of sound in air is,
$\dfrac{{{V_T}}}{{{V_0}}} = \sqrt {\dfrac{T}{{{T_0}}}} $
(Here ${V_T}$ is the velocity of air at temp $T$(in kelvin) and ${V_0}$ is the velocity of air at temperature ${T_0}$(in kelvin))
As given in the question ${V_0}$= 332 m/s and $T$= ${20^0}C$ and ${T_0}$= ${0^0}C$.
Converting temperatures in kelvin
$T$= 20 + 273 = 293K
$\Rightarrow {T_0}$= 0 + 273 =273K
Substituting the given values
$\dfrac{{{V_T}}}{{332}} = \sqrt {\dfrac{{293}}{{273}}} $
$\Rightarrow \dfrac{{{V_T}}}{{332}} = 1.035$
Therefore ${V_T} = 343.62\,m/s$.
Doppler effect: The Doppler Effect is the shift in wave frequency that occurs as a wave source moves relative to its observer. When a sound source approaches you, for example, the frequency of the sound waves changes, resulting in a higher pitch. Applying Doppler effect when whistle is approaching observer
$\dfrac{{{f_a}}}{{{f_0}}} = \dfrac{{{V_T}}}{{{V_T} - {V_W}}}$
When whistle is distancing from observer,
$\dfrac{{{f_d}}}{{{f_0}}} = \dfrac{{{V_T}}}{{{V_T} + {V_W}}}$
Dividing these equations, we get,
$\dfrac{{{f_d}}}{{{f_a}}} = \dfrac{{{V_T} - {V_W}}}{{{V_T} + {V_W}}}$
As given in the question $\dfrac{{{f_d}}}{{{f_a}}} = \dfrac{4}{5}$
$\dfrac{4}{5} = \dfrac{{{V_T} - {V_W}}}{{{V_T} + {V_W}}}$ and by equating further we get
$\Rightarrow {V_W} = \dfrac{{{V_T}}}{9}$
$\therefore {V_W} = 38.2\,m/s$.........(By substituting value of ${V_T} = 343.62\,m/s$)
Hence, the correct answer is option C.
Note:Molecules with more energy can vibrate faster at higher temperatures. Sound waves can fly faster because molecules vibrate faster. In room temperature air, sound travels at 346 metres per second.
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