Question

An ambulance emitting a whine at \(1602 \mathrm{~Hz}\) overtakes and passes a cyclist pedaling a bike at \(2.63 \mathrm{~m} / \mathrm{s}\). After being passed, the cyclist hears a frequency of \(1590 \mathrm{~Hz}\). How fast is the ambulance moving?

Short answer

The ambulance is moving at a speed of approximately \(13.96 \mathrm{~m}/\mathrm{s}\).

Step-by-step solution

Analyze the problem and identify the given data

From the problem, the following data can be extracted: Source frequency (the frequency of ambulance whine), \(f_{\mathrm{s}} = 1602 \mathrm{~Hz}\), Velocity of the observer (speed of the cyclist), \(v_{\mathrm{o}} = 2.63 \mathrm{~m} / \mathrm{s}\), and the observed frequency after the ambulance passes, \(f_{\mathrm{o}}' = 1590 \mathrm{~Hz}\). We are asked to find the velocity of the source (the speed of the ambulance), \(v_{\mathrm{s}}\). The speed of sound in air is \(v_{\mathrm{air}} = 343 \mathrm{~m}/\mathrm{s}\). Note that because the ambulance is moving towards the cyclist before passing and away from the cyclist after passing, we will see an increase in frequency (blue shift) followed by a decrease in frequency (red shift).

Apply Doppler Effect formula for moving source and stationary observer before the ambulance passes the cyclist

The formula for the Doppler Effect when the source is moving towards the observer is: \(f' = f_{\mathrm{s}} \cdot \((v_{\mathrm{air}} + v_{\mathrm{o}})\) / \((v_{\mathrm{air}} - v_{\mathrm{s}})\). Here, \(f'\) is the frequency the cyclist would hear before being overtaken. We can set \(f'\) to be equal to the source frequency \(f_{\mathrm{s}}\) because the problem doesn't specify a change in frequency until after the ambulance passes the cyclist. Thus, \(1602 =1602 \cdot \((343 + 2.63)\) / \((343 - v_{\mathrm{s}})\), which simplifies to (343 - \(v_{\mathrm{s}}\) ) = 345.63.

Apply Doppler Effect formula for moving source and stationary observer after the ambulance passes the cyclist

The formula for the Doppler Effect when the source is moving away from the observer is: \(f'' = f_{\mathrm{s}} \cdot \((v_{\mathrm{air}} - v_{\mathrm{o}})\) / \((v_{\mathrm{air}} + v_{\mathrm{s}})\). Here, \(f'' = 1590 \mathrm{~Hz}\) is the frequency the cyclist hears after being overtaken. Thus, \(1590 = 1602 \cdot \((343 - 2.63)\) / \((343 + v_{\mathrm{s}})\). Multiply both sides by (343 + \(v_{\mathrm{s}}\)) to get \(1590 \cdot (343 + v_{\mathrm{s}}) = 1602 \cdot (340.37)\).

Solve the two equations to find \(v_{\mathrm{s}}\)

We now have two equations: (343 - \(v_{\mathrm{s}}\)) = 345.63 and 1590 \cdot (343 + \(v_{\mathrm{s}})) = 1602 \cdot 340.37. Upon solving this system of equations, find that the speed of the ambulance (\(v_{\mathrm{s}}\)) is approximately 13.96 m/s.

Key concepts

Frequency Shift

The frequency shift is a fascinating concept, often associated with the Doppler Effect. It's a phenomenon where the frequency of a wave changes for an observer moving relative to the source of the wave. In our exercise, this concept is demonstrated by the sound of an ambulance as it approaches and passes by a cyclist.

When the ambulance approaches, the sound waves become compressed, resulting in higher frequency sounds (blue shift). After it passes, the waves are stretched, causing a lower frequency (red shift). This shift is what the cyclist perceives as the change in pitch.

This effect is critical in various wave phenomena, beyond just sound waves, including light and electromagnetic waves. Understanding frequency shifts helps explain everyday experiences, like hearing varying pitches of sirens, and applications in astronomy for measuring distant stars' velocities.

Velocity Calculation

Velocity calculation plays a pivotal role in understanding the movement of objects relative to each other. In our scenario, it helps determine the speed of the ambulance based on the observed frequency changes of its siren sound.

To calculate the velocity, we use the Doppler Effect formulas for sound. When the ambulance is moving towards the observer, the effective observed frequency increases. Conversely, when moving away, the frequency decreases. By setting up equations using the known frequencies and speeds, we solve for the unknown velocity of the ambulance.

This calculation is significant because it showcases practical applications of physics in determining speeds without direct measurement, simply by analyzing sound. This analysis is essential in many fields, including radar and astronomy.

Wave Phenomena

Wave phenomena encompass a vast range of behaviors demonstrated by waves in different mediums. The Doppler Effect is a prime example of wave interaction with moving observers or sources.

Sound waves, as seen with the ambulance scenario, are one type of wave that exhibits classic wave phenomena. These waves propagate through air, and their interaction with moving objects leads to frequency shifts we've discussed. However, other waves, such as light, also experience similar shifts under the influence of motion.

• Sound waves require a medium to travel, illustrating how medium properties affect wave speed.
• Light waves do not require a medium, highlighting differences in wave behavior across types.

Observing these phenomena aids in understanding foundational physics concepts like wave speed, frequency, and the impact of velocity on perception. This understanding is crucial in areas like acoustics, optics, and even quantum physics.