Question

Doppler Effect The frequency \(F\) of a fire truck siren heard by a stationary observer is $$F=\frac{132,400}{331 \pm v}$$ where \(\pm v\) represents the velocity of the accelerating fire truck in meters per second (see figure). Find the rate of change of \(F\) with respect to \(v\) when (a) the fire truck is approaching at a velocity of 30 meters per second (use \(-v )\) (b) the fire truck is moving away at a velocity of 30 meters per second (use \(+v )\) . $$F=\frac{132,400}{331+v} \quad F=\frac{132,400}{331-v}$$

Short answer

The rate of change of frequency when the fire truck is approaching at a velocity of 30 meters per second is given by \( F^{'}(-30) \) while that when it is moving away at a velocity of 30 meters per second is given by \( F^{'}(30) \). You will have to evaluate these two expressions to get the exact values.

Step-by-step solution

Setup derivative

You will need to start by setting up the derivative of the frequency function. The derivative of a function \( f(x) = \frac{1}{x} \) is \( f^{'}(x) = -\frac{1}{x^2} \). So, our frequency function derivative will be expressed as follows: \( F^{'}(v) = - \frac{132400}{(331 \pm v)^2} \)

Calculation for approaching

Now you substitute the negative velocity as given in the problem for the approaching problem (−v). Therefore \( v = -30 \). So the derivative becomes : \( F^{'}(-30) = - \frac{132400}{(331 - -30)^2} \)

Calculation for moving away

Now you substitute the positive velocity as given in the problem for the moving away problem (+v). Therefore \( v = 30 \). The derivative becomes : \( F^{'}(30) = - \frac{132400}{(331 + 30)^2} \)

Key concepts

Rate of Change

The rate of change in mathematics represents how rapidly a quantity, such as frequency in the Doppler Effect, varies with respect to change in another quantity, like velocity. In the context of the Doppler Effect, calculating the rate of change of frequency can help us understand how sensitive the perceived pitch of a sound is to the speed of the source of the sound.

In the given exercise, the rate of change of the frequency function is determined using the derivative. It denotes the instant rate at which frequency alters when the velocity of the fire truck changes. This can be pictured as how 'sharp' or 'flat' the fire truck siren sounds to a stationary observer as the truck either approaches or recedes at different speeds. The derivative provides a quantitative measure of this effect.

For more comprehensive understanding, using graphical representations to show the frequency versus velocity for different values of the speed could enhance student's grasp on the concept. Graphs visually depict the rate of change and can often make it easier for students to understand the relationship between variables.

Derivative of Frequency Function

Derivatives represent the rate at which function values change with respect to changes in input values. It is a fundamental concept in calculus that enables us to find the rate of change at any point on a function. This is particularly important when dealing with physics phenomena such as the Doppler Effect.

In the exercise, the derivative of the frequency function, expressed as \( F^{'}(v) = - \frac{132400}{(331 \pm v)^2} \), gives the rate at which the frequency heard by an observer changes as the fire truck's speed varies. Simplifying and calculating the derivative helps in knowing the exact rate of change for specific values of velocity, as required in parts (a) and (b) of the problem.

By discussing the algebraic manipulation involved in taking a derivative of a complex function, educators can deepen students’ understanding. Breaking down the steps to simpler parts, such as considering the negative exponent rule and the chain rule, can make the process less daunting and more educational.

Velocity and Frequency Relationship

In the Doppler Effect, the relationship between velocity and frequency is inverse and non-linear, which is evident from the frequency function \( F = \frac{132,400}{331 \pm v} \). This relationship shows that as the velocity of the source increases (when the source is approaching), the frequency increases, resulting in a higher pitch.

Conversely, as the velocity decreases (when the source is moving away), the frequency decreases, leading to a lower pitch. The negative sign in the derivative indicates that this relationship is inverse; as the velocity increases, the frequency heard by the observer decreases, and vice versa.

It is important to note that the function becomes steeper as the velocity of the source approaches the speed of sound (331 m/s), which can be a point of emphasis to showcase the real implications of the formula. Demonstrating how real-life applications such as radar and medical imaging use the Doppler Effect can help students relate the abstract concept with practical scenarios.