Question

On a quiet day, when a whistle crosses you then its pitch decreases in the ratio $⅘$. If the temperature on that day is ${20^0}C$, then the speed of whistle will be ____ (given the velocity of sound at ${0^0}= 332\,m/s$)
A. 48.2 m/s
B. 68.2 m/s
C. 38.2 m/s
D. 108.2 m/s

Answer 1

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.