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A wireless transmitting microphone is mounted on a small platform that can roll down an incline, directly away from a loudspeaker that is mounted at the top of the incline. The loudspeaker broadcasts a tone that has a fixed frequency of \(1.000 \times 10^{4} \mathrm{Hz},\) and the speed of sound is \(343 \mathrm{m} / \mathrm{s} .\) At a time of \(1.5 \mathrm{s}\) following the release of the platform, the microphone detects a frequency of \(9939 \mathrm{Hz}\). At a time of \(3.5 \mathrm{s}\) following the release of the platform, the microphone detects a frequency of \(9857 \mathrm{Hz}\) What is the acceleration (assumed constant) of the platform?

Short Answer

Expert verified
The acceleration of the platform is approximately \(1.82 \, \mathrm{m/s}^2\).

Step by step solution

01

Understanding the Problem

We have a source frequency, a platform moving away, and detected frequencies at specific times. The problem asks for the acceleration of the platform, knowing it rolls directly away from the sound source (Doppler effect application).
02

Recall the Doppler Effect Formula

For a source moving away from an observer at speed \(v_s\), the observed frequency \(f'\) is given by: \[f' = \frac{f}{1 + \frac{v_s}{v}}\]where \(f\) is the emitted frequency, \(v_s\) is the speed of the source, and \(v\) is the speed of sound.
03

Setup Equation for First Time Point

At \(t = 1.5\, \mathrm{s}\), the detected frequency is \(9939\, \mathrm{Hz}\). Substitute into the Doppler equation: \[9939 = \frac{10000}{1 + \frac{v_{s,1}}{343}}\]Solve for \(v_{s,1}\) (speed at \(t = 1.5\, \mathrm{s}\)).
04

Solve for Speed at First Time Point

Rearrange the equation to solve for \(v_{s,1}\):\[1 + \frac{v_{s,1}}{343} = \frac{10000}{9939}\] \[\frac{v_{s,1}}{343} = \frac{10000}{9939} - 1\]\[v_{s,1} \approx 3.46\, \mathrm{m/s}\]
05

Setup Equation for Second Time Point

At \(t = 3.5\, \mathrm{s}\), the detected frequency is \(9857\, \mathrm{Hz}\). Substitute into the Doppler equation:\[9857 = \frac{10000}{1 + \frac{v_{s,2}}{343}}\]Solve for \(v_{s,2}\) (speed at \(t = 3.5\, \mathrm{s}\)).
06

Solve for Speed at Second Time Point

Rearrange the equation to solve for \(v_{s,2}\):\[1 + \frac{v_{s,2}}{343} = \frac{10000}{9857}\]\[\frac{v_{s,2}}{343} = \frac{10000}{9857} - 1\]\[v_{s,2} \approx 7.10\, \mathrm{m/s}\]
07

Determine the Acceleration of the Platform

The platform's speed increases from \(v_{s,1}\) to \(v_{s,2}\). Use the formula for acceleration:\[a = \frac{v_{s,2} - v_{s,1}}{t_2 - t_1}\]\[a = \frac{7.10 - 3.46}{3.5 - 1.5}\]\[a \approx 1.82\, \mathrm{m/s}^2\]
08

Verify the Solution

Re-calculate using the derived acceleration value to check consistency. Given the detected frequencies align with the calculated speeds and subsequent acceleration, the steps and solution check out.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sound Frequency Perception
When studying sound waves and their perception, we delve into how frequencies are altered by motion. The Doppler Effect is the cornerstone concept here. It explains why a sound changes pitch as a source moves relative to an observer. Imagine a sound wave as ripples on a pond. If the source of the ripple moves, the ripples in front of it compress (higher frequency or pitch) while those behind spread apart (lower frequency or pitch).

In our exercise, a sound frequency emitted at 10,000 Hz is detected at a lower frequency by the moving microphone. This shift occurs because the microphone moves away from the sound source, encountering fewer wavefronts per second.

The change in pitch you hear as a car passes by is a classic everyday example of the Doppler Effect. The faster the object moves, the greater the frequency shift. Understanding these changes helps you solve questions about motion and sound perception in various physics problems.
Constant Acceleration
Constant acceleration happens when an object's velocity changes at a steady rate over time. In our problem, the platform moves down the incline with constant acceleration, increasingly speeding up as it goes. This smooth acceleration results in an increased speed at different moments, as seen from the two distinct frequencies recorded.

To find acceleration, we can use the formula:
  • \( a = \frac{v_{final} - v_{initial}}{t_{final} - t_{initial}} \)
This equation calculates how speed changes over time. Here, the Doppler Effect provides the platform's speed at given times by relating frequency shifts back to velocity.

Understanding constant acceleration is crucial in predicting an object's future motion. You use this concept to extrapolate speeds and positions from past data, a key skill in physics and engineering.
Speed of Sound
The speed of sound is the rate at which sound waves propagate through a medium. It significantly affects how we perceive frequency changes through the Doppler Effect. In our case, the speed of sound is given as 343 m/s, which is typical for sound waves traveling through air at room temperature.

This speed varies based on factors like temperature, air pressure, and medium. For instance, sound travels faster in water and metals than in air due to the density and elasticity of these materials.

Knowing the speed of sound allows us to accurately calculate how much the frequency changes as our platform moves. The formula used in these calculations was:
  • \( f' = \frac{f}{1 + \frac{v_s}{v}} \)
where \(v\) is the speed of sound. This context is essential for interpreting how frequency shifts relate to relative motion and solving problems around sound and acceleration.

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Most popular questions from this chapter

From a vantage point very close to the track at a stock car race, you hear the sound emitted by a moving car. You detect a frequency that is 0.86 times as small as the frequency emitted by the car when it is stationary. The speed of sound is \(343 \mathrm{m} / \mathrm{s} .\) What is the speed of the car?

Two wires are parallel, and one is directly above the other. Each has a length of \(50.0 \mathrm{m}\) and a mass per unit length of \(0.020 \mathrm{kg} / \mathrm{m} .\) However, the tension in wire \(\mathrm{A}\) is \(6.00 \times 10^{2} \mathrm{N},\) and the tension in wire \(\mathrm{B}\) is \(3.00 \times 10^{2} \mathrm{N} .\) Transverse wave pulses are generated simultaneously, one at the left end of wire \(\mathrm{A}\) and one at the right end of wire \(\mathrm{B}\). The pulses travel toward each other. How much time does it take until the pulses pass each other?

A typical adult ear has a surface area of \(2.1 \times 10^{-3} \mathrm{m}^{2}\). The sound intensity during a normal conversation is about \(3.2 \times 10^{-6} \mathrm{W} / \mathrm{m}^{2}\) at the listener's ear. Assume that the sound strikes the surface of the ear perpendicularly. How much power is intercepted by the ear?

A wire is stretched between two posts. Another wire is stretched between two posts that are twice as far apart. The tension in the wires is the same, and they have the same mass. A transverse wave travels on the shorter wire with a speed of \(240 \mathrm{m} / \mathrm{s}\). What would be the speed of the wave on the longer wire?

A source of sound is located at the center of two concentric spheres, parts of which are shown in the drawing. The source emits sound uniformly in all directions. On the spheres are drawn three small patches that may or may not have equal areas. However, the same sound power passes through each patch. The source produces \(2.3 \mathrm{W}\) of sound power, and the radii of the concentric spheres are \(r_{A}=0.60 \mathrm{m}\) and \(r_{\mathrm{B}}=0.80 \mathrm{m}\) (a) Determine the sound intensity at each of the three patches. (b) Thesound power that passes through each of the patches is \(1.8 \times 10^{-3} \mathrm{W}\). Find the area of each patch.

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