
A sound wave has a frequency of 192 Hz and travels the length of the football field of $ 91.4m $ in $ 0.27s $ . What is the speed of the wave?
(A) $ 338.5m/s $
(B) $ 348.5m/s $
(C) $ 358.5m/s $
(D) $ 368.5m/s $
Answer
582.6k+ views
Hint To solve this question, we need to use the basic formula for the velocity of an object and thus obtain the value of velocity of the wave.
Formula Used: The formula used to solve this question is
$\Rightarrow v = \dfrac{x}{t} $
Here $ v $ is the velocity of the object, $ x $ is the distance travelled, and $ t $ is the time taken.
Complete step by step answer
Let the distance covered by the sound wave be $ x $ and the time taken be $ t $
According to the question, the wave covers a displacement of $ 91.4m $ in $ 0.27s $ time. So we have
$\Rightarrow x = 91.4m $
$\Rightarrow t = 0.27s $
We know from the basic definition of the velocity that
$\Rightarrow v = \dfrac{x}{t} $
Substituting from above, we get
$\Rightarrow v = \dfrac{{91.4}}{{0.27}} $
On solving we get
$\Rightarrow v = 338.5m/s $
$ \therefore $ The correct answer is option A.
Additional Information:
We know that the speed of sound waves is $ 320m/s $ . But the value speed of waves that we’ve got is greater than the speed of the sound in air at NTP. Thus we can conclude that either the wave travels in a rarer medium or the conditions are not at NTP. For example, the speed of sound waves is greater when the pressure drops below $ 1{\text{ b}}ar $ during the typhoons and hurricanes or when the temperature increases from $ \;300K $ during summer time. Thus, speed of sound can vary and is not constant unlike the general opinion. But we should take the speed of sound to be $ 320m/s $ if its value is not provided in the question.
Note
This is a simple question where the data related to frequency of sound is given just to confuse us. In no case should we use frequency and take its reciprocal to calculate the time and put in the formula for velocity. The reason for this is that the reciprocal of frequency gives us the time period of the wave and not the time taken to travel a given distance.
Formula Used: The formula used to solve this question is
$\Rightarrow v = \dfrac{x}{t} $
Here $ v $ is the velocity of the object, $ x $ is the distance travelled, and $ t $ is the time taken.
Complete step by step answer
Let the distance covered by the sound wave be $ x $ and the time taken be $ t $
According to the question, the wave covers a displacement of $ 91.4m $ in $ 0.27s $ time. So we have
$\Rightarrow x = 91.4m $
$\Rightarrow t = 0.27s $
We know from the basic definition of the velocity that
$\Rightarrow v = \dfrac{x}{t} $
Substituting from above, we get
$\Rightarrow v = \dfrac{{91.4}}{{0.27}} $
On solving we get
$\Rightarrow v = 338.5m/s $
$ \therefore $ The correct answer is option A.
Additional Information:
We know that the speed of sound waves is $ 320m/s $ . But the value speed of waves that we’ve got is greater than the speed of the sound in air at NTP. Thus we can conclude that either the wave travels in a rarer medium or the conditions are not at NTP. For example, the speed of sound waves is greater when the pressure drops below $ 1{\text{ b}}ar $ during the typhoons and hurricanes or when the temperature increases from $ \;300K $ during summer time. Thus, speed of sound can vary and is not constant unlike the general opinion. But we should take the speed of sound to be $ 320m/s $ if its value is not provided in the question.
Note
This is a simple question where the data related to frequency of sound is given just to confuse us. In no case should we use frequency and take its reciprocal to calculate the time and put in the formula for velocity. The reason for this is that the reciprocal of frequency gives us the time period of the wave and not the time taken to travel a given distance.
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