
A bat moving at \[10m{s^{ - 1}}\] towards a wall sends a sound signal of $8000Hz$ towards it. On reflection, it hears a sound of frequency f. The value of f in Hz is close to (speed of sound=$320m{s^{ - 1}}$ )
A) $8258$
B) $8424$
C) $8000$
D) $8516$
Answer
573.6k+ views
Hint: Sound waves are the waves that travel at a certain speed and has wavelength and frequency. Frequency is defined as the number of occurrences in a unit time. In the case of a sound wave, the frequency of a sound wave is defined as the rate of vibration of the sound waves through the air.
Complete step by step answer:
Step I:
When a sound wave strikes a surface, then it reflects back from the surface. The angle of incidence at which the sound wave strikes is equal to the angle of reflection.
Step II:
The time taken by a sound wave to travel a distance S towards the wall can be calculated using the formula:
$S = v.t$
Or $t = \dfrac{s}{v}$
Where S is the displacement
V is the velocity and
T is the time taken
Step III:
$t = \dfrac{s}{{{v_{sound}} - {v_{bat}}}}$
It is known that frequency varies inversely with time.
$\Rightarrow t = \dfrac{1}{f} = \dfrac{1}{{8000}}$
$\Rightarrow \dfrac{s}{{{v_{sound}} - {v_{bat}}}} = \dfrac{1}{{8000}}$
$\Rightarrow s = \dfrac{{{v_{sound}} - {v_{bat}}}}{{8000}}$ -------(i)
Step IV:
Given that the bat is moving towards the wall, let the time taken by the sound wave to reach the wall is $T'$.
\[T' = \dfrac{{\dfrac{{{v_{sound}} - {v_{bat}}}}{{8000}}}}{{{v_{sound}} + {v_{bat}}}}\] ---------(ii)
Step V:
Now the frequency varies inversely with the time period, therefore the frequency of the sound wave on reflection is given by
$\Rightarrow F' = \dfrac{1}{{T'}}$
$\Rightarrow F' = \dfrac{{{v_{sound}} + {v_{bat}}}}{{{v_{sound}} - {v_{bat}}}} \times 8000$
$\Rightarrow F' = \dfrac{{(10 + 320)8000}}{{320 - 10}}$
$\Rightarrow F' = 8516Hz$
The frequency of the sound wave after reflection is given by $8516Hz$. Hence, Option (D) is the right answer.
Note:
It is to be noted that whenever there is a change of frequency in the waves of an object moving then it can be calculated using Doppler’s effect. As per this effect, the frequency of the sound waves increases, if the object moves towards the observer. The frequency of the sound wave decreases as the object moves away from the observer.
Complete step by step answer:
Step I:
When a sound wave strikes a surface, then it reflects back from the surface. The angle of incidence at which the sound wave strikes is equal to the angle of reflection.
Step II:
The time taken by a sound wave to travel a distance S towards the wall can be calculated using the formula:
$S = v.t$
Or $t = \dfrac{s}{v}$
Where S is the displacement
V is the velocity and
T is the time taken
Step III:
$t = \dfrac{s}{{{v_{sound}} - {v_{bat}}}}$
It is known that frequency varies inversely with time.
$\Rightarrow t = \dfrac{1}{f} = \dfrac{1}{{8000}}$
$\Rightarrow \dfrac{s}{{{v_{sound}} - {v_{bat}}}} = \dfrac{1}{{8000}}$
$\Rightarrow s = \dfrac{{{v_{sound}} - {v_{bat}}}}{{8000}}$ -------(i)
Step IV:
Given that the bat is moving towards the wall, let the time taken by the sound wave to reach the wall is $T'$.
\[T' = \dfrac{{\dfrac{{{v_{sound}} - {v_{bat}}}}{{8000}}}}{{{v_{sound}} + {v_{bat}}}}\] ---------(ii)
Step V:
Now the frequency varies inversely with the time period, therefore the frequency of the sound wave on reflection is given by
$\Rightarrow F' = \dfrac{1}{{T'}}$
$\Rightarrow F' = \dfrac{{{v_{sound}} + {v_{bat}}}}{{{v_{sound}} - {v_{bat}}}} \times 8000$
$\Rightarrow F' = \dfrac{{(10 + 320)8000}}{{320 - 10}}$
$\Rightarrow F' = 8516Hz$
The frequency of the sound wave after reflection is given by $8516Hz$. Hence, Option (D) is the right answer.
Note:
It is to be noted that whenever there is a change of frequency in the waves of an object moving then it can be calculated using Doppler’s effect. As per this effect, the frequency of the sound waves increases, if the object moves towards the observer. The frequency of the sound wave decreases as the object moves away from the observer.
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