
A racing car passes the stand of a race course traveling at a speed of $ 144\,km/hr $. If the noise from the exhaust has a frequency of $ 300Hz $ , determine the frequency heard by the audience as the car approaches stand. [Assume speed of sound as $ 340m/s $ ]
Answer
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Hint: Here in this question we have to find the frequency heard by an audience, so for this, we will use the concept of doppler's effect, and the formula for it is given by $ {f_s}\dfrac{V}{{V - {V_s}}} $ . So by using this and substituting the values, we will get to the answer.
Formula used:
Doppler’s formula when the source is moving towards the observer and is at rest,
$ {f_0} = {f_s}\dfrac{V}{{V - {V_s}}} $
Here,
$ {f_0} $ , will be the observer frequency
$ {f_s} $ , will be the actual frequency of sound waves
$ V $ , will be the velocity of sound
$ {V_s} $ , will be the source velocity.
Complete step by step solution:
Since the source value is given in kilometers per hour so first of all, we will change the unit of it. So it will be written as
$ \Rightarrow {V_s} = 144 \times \dfrac{5}{{18}}m/s $
And on solving the above equation, we get the equation as
$ \Rightarrow {V_s} = 40m/s $
So now by using the Doppler’s formula and substituting the known values. We will get the equation as
$ \Rightarrow {f_0} = 300\left( {\dfrac{{340}}{{340 - 40}}} \right) $
Now on solving the above equation, we will get the equation as
$ \Rightarrow {f_0} = 300\left( {\dfrac{{340}}{{300}}} \right) $
Now on solving the numerator and denominator part, we will get the equation as
$ \Rightarrow {f_0} = 340Hz $
Therefore, the frequency heard by the audience as the car approaches the stand is equal to $ 340Hz $ .
Note:
Here in this question we have to take care of the units as it becomes an important step and if we don’t do that then our answer will be incorrect. Also, the best example of this effect is one of an ambulance passing by. The frequency of the Siren increases as it comes near us, and then diminishes as it passes away. But if we ask a person inside the Ambulance, the frequency of sound remains the same.
Formula used:
Doppler’s formula when the source is moving towards the observer and is at rest,
$ {f_0} = {f_s}\dfrac{V}{{V - {V_s}}} $
Here,
$ {f_0} $ , will be the observer frequency
$ {f_s} $ , will be the actual frequency of sound waves
$ V $ , will be the velocity of sound
$ {V_s} $ , will be the source velocity.
Complete step by step solution:
Since the source value is given in kilometers per hour so first of all, we will change the unit of it. So it will be written as
$ \Rightarrow {V_s} = 144 \times \dfrac{5}{{18}}m/s $
And on solving the above equation, we get the equation as
$ \Rightarrow {V_s} = 40m/s $
So now by using the Doppler’s formula and substituting the known values. We will get the equation as
$ \Rightarrow {f_0} = 300\left( {\dfrac{{340}}{{340 - 40}}} \right) $
Now on solving the above equation, we will get the equation as
$ \Rightarrow {f_0} = 300\left( {\dfrac{{340}}{{300}}} \right) $
Now on solving the numerator and denominator part, we will get the equation as
$ \Rightarrow {f_0} = 340Hz $
Therefore, the frequency heard by the audience as the car approaches the stand is equal to $ 340Hz $ .
Note:
Here in this question we have to take care of the units as it becomes an important step and if we don’t do that then our answer will be incorrect. Also, the best example of this effect is one of an ambulance passing by. The frequency of the Siren increases as it comes near us, and then diminishes as it passes away. But if we ask a person inside the Ambulance, the frequency of sound remains the same.
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