Question

7.
A person going away from a factory on his scooter at a speed of 36 km/hr listens to
the siren of the factory. If the frequency of siren is 525 Hz and a wind is blowing
along the direction of scooter at 36 km/hr, the frequency heard by the person is
(velocity of sound = 340 m/s)
1) 680 Hz
2)510 Hz
3) 640 Hz
4) 600 Hz

Answer 1

Answer:

Let distance between Manipur Dispur = x km

Average speed of train from Manipur \( \Large = \frac{x}{4} km/h \)

Average speed of train from Dispur \( \Large = \frac{2x}{7} km/h \)

Let they meet y h after 6 : 00 am.

Then, according to the question,

\( \Large \left(\frac{x}{4} \times y\right)+\frac{2x}{7} \times \left(y-2\right) = x \)

\( \Large \frac{y}{4}+\frac{2 \left(y-2\right) }{7} = 1 \)

7y+8(y-2) = 28

15y = 44

y = \( \Large 15y = 44 \) = 2 h 56 min

Clearly, trains meet 2 h 56 min after 6 : 00 am.

It means the trains meet at 8 : 56 am

Answer 2

Given:

Speed of scooter $\[{\[{{v_o}}\] = 36\,km/hr\]$

Frequency of siren, $\[{f_o} = 525\,Hz\]$ $\[ = 525\,Hz\]$

Speed of wind, $\[{v_w} = 36km/hr\]$

To Find: Frequency heard by the person

Solution:

Doppler's effect is observed when a moving person hears a sound from a distance.

The actual frequency of the sound is different from the frequency of the sound heard by the person.

Applying Doppler's effect formula, i.e.,

$\[{f_o}' = {f_o}\frac{{(v + {v_o})}}{{(v - {v_s})}}\]$

where, $\[{f_o}\]=$ actual frequency of sound, $\[{{v_o}}\]=$ speed of observer, $\[{{v_s}}\]=$ speed of source, $v=$ speed of sound

For the given condition the formula becomes

$\[{f_o}' = {f_o}\left( {\frac{{{v_s} + {v_w} - {v_o}}}{{{v_s} + {v_w}}}} \right)\]$                     ... (i)

At first, convert the speed of the scooter into meter/second.

$\[\begin{gathered} {v_o} = 36 \times \frac{5}{{18}} \hfill \\ {v_o} = 10\,m/s \hfill \\ \end{gathered} \]$

The speed of the wind will also be $10 m/s$.

Substituting the values in the equation (i), we have,

$\[\begin{gathered} {f_o}' = 525 \times \left( {\frac{{340 + 10 - 10}}{{340 + 10}}} \right) \hfill \\ {f_o}' = 525 \times \left( {\frac{{340}}{{350}}} \right) \hfill \\ {f_o}' = 510\,Hz \hfill \\ \end{gathered} \]$

Hence, the frequency heard by the person is $\[510\,Hz\]$.

#SPJ2