Question

You and a friend are sitting in the outfield bleachers of a major league baseball park, 140 m from home plate on a day when the temperature is \(20^{\circ} \mathrm{C}\). Your friend is listening to the radio commentary with headphones while watching. The broadcast network has a microphone located \(17 \mathrm{m}\) from home plate to pick up the sound as the bat hits the ball. This sound is transferred as an EM wave a distance of \(75,000 \mathrm{km}\) by satellite from the ball park to the radio. (a) When the batter hits a hard line drive, who will hear the "crack" of the bat first, you or your friend, and what is the shortest time interval between the bat hitting the ball and one of you hearing the sound? (b) How much later does the other person hear the sound?

Short answer

Your friend hears the sound first. The shortest time is 0.2996 s; you hear it 0.1084 s later.

Step-by-step solution

Calculate Speed of Sound

The speed of sound in air at \(20^{\circ} \mathrm{C}\) is approximately \(343 \mathrm{m/s}\). This will help us determine how long it takes for sound to travel from home plate to the listener.

Calculate the Distance Sound Travels to the Fan

You, the fan sitting in the bleachers, are \(140 \mathrm{m}\) from home plate. Calculate the time \(t_1\) taken for the sound to travel this distance using the formula: \[ t_1 = \frac{\text{Distance}}{\text{Speed}} = \frac{140 \mathrm{m}}{343 \mathrm{m/s}} \approx 0.408 \text{ s} \]

Calculate the Total Distance Sound Travels to Your Friend

The microphone is \(17 \mathrm{m}\) from home plate. The sound first travels \(17 \mathrm{m}\) to the microphone and then \(75,000,000 \mathrm{m}\) (or \(75,000\, \, \mathrm{km}\)) via satellite. Calculate the EM sound wave distance time \(t_2\): \[ t_2 = \frac{17 \mathrm{m}}{343 \mathrm{m/s}} + \frac{75,000,000 \mathrm{m}}{3 \times 10^8 \mathrm{m/s}} \]The first part of the journey takes approximately \(0.0496 \text{ s}\), and the second part is approximately \(0.25 \text{ s}\), giving \[t_2 \approx 0.2996 \text{ s}\].

Compare Time Intervals

Compare \(t_1\) and \(t_2\). You, the fan, hear the sound first at \(0.408 \text{ s}\) after the crack, while your friend hears it via broadcast at \(0.2996 \text{ s}\). Thus, the shortest interval for hearing the sound is for your friend at \(0.2996\) seconds.

Calculate the Delay for the Fan

The delay experienced by you, the fan, can be calculated by subtracting the two times: \[ 0.408 \text{ s} - 0.2996 \text{ s} = 0.1084 \text{ s}\]Thus, you hear the sound \(0.1084\) seconds later.

Key concepts

Speed of Sound

The speed of sound is the rate at which sound waves travel through a medium. At 20^{\circ} \mathrm{C}, the speed in dry air is around 343 meters per second (\mathrm{m/s}). This speed is not constant and can vary with temperature and pressure.

• Sound is a mechanical wave, meaning it needs a medium like air, water, or solids to travel through.
• The molecules in the medium vibrate, carrying the wave from one location to another.
• In general, sound travels faster in solids, then liquids, and slowest in gases due to the density of particles.

Understanding the speed of sound is crucial when calculating how long it takes for a sound to travel a given distance. We often use the formula: \[ \text{Time} = \frac{\text{Distance}}{\text{Speed of Sound}} \] to find travel time.

Electromagnetic Waves

Electromagnetic (EM) waves are crucial for transmitting information over long distances, like radio broadcasts. Unlike sound waves, EM waves do not require a medium and can travel through the vacuum of space. This makes them ideal for satellite communication.

• EM waves move at the speed of light, which is approximately \(3 \times 10^8 \, \mathrm{m/s}\).
• The fact that they don't need a medium allows them to cover vast distances (such as the 75,000 km to a satellite) almost instantaneously.
• Satellites and antennas receive these waves, convert them to signals, and disseminate them for public use such as radio listening.

In physics problems, the large speed of EM waves plays a key role in calculationslike comparing how sound travels to different points contrastingly faster than EM waves handle broadcast signals.

Distance and Time Calculations

In problems that involve sound or light travel, calculating distance and timeis a foundational skill. Distances can greatly influence how soon sounds areheard or signals are received.

• Understanding distances in a setup helps determine which parts require the speed of sound or the speed of light (EM waves) in calculations.
• For sound: using \[ \text{Time} (t_1) = \frac{140 \mathrm{m}}{343 \, \mathrm{m/s}} \], as shown, determines the travel time to the fan.
• For the sound wave picked up by a microphone, the travel is broken into two:17 m by sound to the microphone and 75,000 km via a satellite at EM wave speed.

This dual calculation method allows us to compare the two distinct pathways ofthe sound as presented in our problem, illustrating how each follows uniqueconditions depending on the medium.

Physics Problem Solving

Physics problem-solving often begins by clearly defining what is given and whatneeds solving. With the baseball scenario, each element (fan location, friend with radio) requires understanding of physical principles to reach a solution.

• Organize data: Clearly list known values — distances, mediums, speeds — before calculations.
• Apply appropriate formulas: Use correct formulations for sound versus EM waves.
• Interpret results: Compare times and decide which event happens first based on numerical outputs.

Problem-solving also benefits from practice in breaking issues into steps as a way of methodically deducing time sequences and outcomes, just like comparing times \(t_1\) and \(t_2\) in the example.