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Two eagles fly directly toward one another, the larger eagle at a speed of 15.0 metres per second and the smaller one at a speed of 20.0 metres per second.
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Both eagles screech, the larger emitting a screech of frequency of 3200 hertz and the smaller one emitting a screech of frequency 3800 hertz.
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The speed of sound in the air around the eagles is 330 metres per second.
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What is the frequency of the screech heard by the larger eagle?
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What is the frequency of the screech heard by the smaller eagle?
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Weβre told in the statement that the larger eagle is moving at a speed of 15.0 metres per second and emits a screech of frequency of 3200 hertz.
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The smaller eagle is moving at 20.0 metres per second and gives off a sound of frequency of 3800 hertz.
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The speed of sound in air is 330 metres per second.
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We want to know the frequency of the screech heard by the larger eagle, what weβll call π sub LH, and we also want to know the frequency of the screech heard by the smaller eagle, what weβll call π sub SH.
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Letβs begin our solution by drawing a diagram.
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We have two eagles flying right at one another; the larger one moving at a speed π£ sub πΏ equals 15.0 metres per second and the smaller one moving at a speed of π£ sub π equals 20.0 metres per second.
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As they fly, the eagles screech, each with its own frequency.
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The smaller eagle screeches at 3800 hertz and the larger one screeches at a frequency π sub πΏ of 3200 hertz.
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We know the observed frequencies of each eagle will be shifted because of the fact that both are in motion.
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This is called a Doppler shift, which is described by the Doppler equation, which says that the observed frequency π sub π equals the speed of sound π£ plus the speed of the observer π£ sub π divided by the speed of sound minus the speed of the source π£ sub π all multiplied by π sub π , the original source frequency.
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When we apply this relationship to our scenario, solving first for the frequency observed by the larger eagle, that frequency equals the speed of sound plus the speed of the larger eagle π£ sub πΏ divided by the speed of sound minus the speed of the smaller eagle π£ sub π all multiplied by the source frequency from the smaller eagle π sub π .
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In this problem, weβll treat the speed of sound π£ as exactly 330 metres per second.
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Knowing that, we can now plug in the variables in this equation to solve for π sub LH.
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When we plug in for these values, we see that the numerator and denominator of our fraction simplify.
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And when we compute this result, we find that π sub LH is equal to 4.23 kilohertz; thatβs the frequency the larger eagle hears from the smaller eagle.
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Now, we move onto solving for π sub SH, the frequency observed by the smaller eagle.
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In this case, since the smaller eagle is the observer, our equation for the frequency shift reads π£ plus π£ sub π divided by π£ minus π£ sub πΏ all multiplied by the frequency the larger eagle emits, π sub πΏ.
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When we plug in for these values to our equation, we again see that the numerator and denominator of the fraction simplify.
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When we calculate this value, we find π sub SH is 3.56 kilohertz; thatβs the frequency the smaller eagle observes coming from the larger one.