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At the Winter Olympics, an athlete rides her luge down the track while a bell just above the wall of the chute rings continuously. When her sled passes the bell, she hears the frequency of the bell fall by the musical interval called a minor third. That is, the frequency she hears drops to five sixths of its original value. (a) Find the speed of sound in air at the ambient temperature \(-10.0^{\circ} \mathrm{C} .\) (b) Find the speed of the athlete.

Short Answer

Expert verified
The speed of sound in air at -10.0°C is approximately 325 m/s, and the speed of the athlete on the luge sled is around 54.2 m/s.

Step by step solution

01

Find the Speed of Sound in Air at -10.0°C

The speed of sound in air varies directly with the square root of the absolute temperature (in Kelvin). The standard speed of sound at 0°C is approximately 331 m/s. The formula to adjust this speed for other temperatures is \( v = v_{0} \sqrt{\frac {T}{T_{0}}} \), where \( v_{0} \) is the speed of sound at 0°C, \( T\) is the absolute temperature and \( T_{0} \) is the standard temperature in Kelvin. First, convert -10.0°C to Kelvin by adding 273.15. Plug in the values into the equation and solve for \( v \).
02

Determine the Frequency Ratio

A minor third musical interval corresponds to a frequency ratio of 5:6. This means the observed frequency after the slight passes is 5/6 of the original frequency.
03

Find the Speed of the Athlete

The speed of the observer (in this case, the athlete on a sled) can be calculated from the formula for the Doppler effect: \( v_{o} = \frac{v(1 - \frac{f}{f_{0}})}{1 - \frac{f}{f_{0}}} \). Here, \( f_{0} \) is the original frequency, \( f \) is the observed frequency (which is 5/6 of the original frequency), \( v \) is the speed of sound, and \( v_{o} \) is the speed of the observer (in this case, the speed of the sled). Plug in the values and solve for \( v_{o} \).

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Key Concepts

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

Speed of Sound
The speed of sound is affected by the medium it travels through, and also by the medium's temperature. In air, sound travels because air molecules collide with each other, transferring energy from one molecule to the next. The speed of sound at 0°C in dry air is approximately 331 meters per second.

To adjust for different temperatures, we use a formula that involves the temperature in Kelvin:
  • Convert Celsius to Kelvin by adding 273.15 to the Celsius temperature.
The formula to find the speed of sound at a particular temperature is:
  • \[ v = v_{0} \sqrt{\frac {T}{T_{0}}} \]
Here, \(v_{0}\) is the speed of sound at 0°C, \(T\) represents the absolute temperature in Kelvin, and \(T_{0}\) is the reference temperature at 0°C which is 273.15 K.

For example, at -10.0°C, you must first convert it to Kelvin, resulting in 263.15 K, and use the formula to find that the speed of sound is about 325 m/s.
Temperature Conversion
Understanding and converting temperature units is essential in physics, especially when dealing with equations that require specific units. The most common units are Celsius, Kelvin, and Fahrenheit.

For the speed of sound calculations, temperature must be in Kelvin:
  • Convert Celsius to Kelvin by adding 273.15 to the Celsius reading: \( K = °C + 273.15\)
This conversion is crucial because Kelvin is the SI unit for temperature and is used in many scientific calculations to ensure accuracy.
If you ever need to convert Celsius to Fahrenheit, the formula is:
  • \[ ^{\circ}F = ^{\circ}C \times \frac{9}{5} + 32 \]
Though not required for this specific exercise, it's good to know for general understanding.
Frequency Ratio
In music and sound, a frequency ratio is important in defining musical intervals. It describes how frequencies relate to each other. For instance, the interval of a minor third has a frequency ratio of 5:6.

This means if a bell rings at a certain frequency and you move so that you hear a minor third lower, the sound you hear will be 5/6 of the original frequency.
  • Original frequency (\(f_0\)) and heard frequency (\(f\)) are related by the ratio \( \frac{f}{f_0} = \frac{5}{6} \).
This concept relates strongly to the Doppler effect, where the observed frequency changes because of motion relative to the sound source.
Musical Intervals
Musical intervals are the difference in pitch between two sounds. They are analogous to ratios in frequency. Different intervals are perceived as different notes that fit into scales and chords.

The minor third, a specific musical interval, is known for its emotive and sometimes melancholic sound. It involves a frequency ratio of 5:6.
  • This means the frequency of the lower note is 5/6 of the frequency of the higher note.
Intervals like the minor third help explain why certain patterns of notes in music are pleasing or recognizable.
The knowledge of intervals is also useful in real-world physics situations, such as analyzing the Doppler effect in this exercise, where the perceived frequency changed by this exact interval due to the athlete passing by the bell.

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Most popular questions from this chapter

A tuning fork vibrating at \(512 \mathrm{Hz}\) falls from rest and accelerates at \(9.80 \mathrm{m} / \mathrm{s}^{2} .\) How far below the point of \(\mathrm{re}-\) lease is the tuning fork when waves of frequency \(485 \mathrm{Hz}\) reach the release point? Take the speed of sound in air to be \(340 \mathrm{m} / \mathrm{s}\)

As a certain sound wave travels through the air, it produces pressure variations (above and below atmospheric pressure) given by \(\Delta P=1.27 \sin (\pi x-340 \pi t)\) in SI units. Find (a) the amplitude of the pressure variations, (b) the frequency, (c) the wavelength in air, and (d) the speed of the sound wave.

When high-energy charged particles move through a transparent medium with a speed greater than the speed of light in that medium, a shock wave, or bow wave, of light is produced. This phenomenon is called the Cerenkov effect. When a nuclear reactor is shielded by a large pool of water, Cerenkov radiation can be seen as a blue glow in the vicinity of the reactor core, due to high-speed electrons moving through the water. In a particular case, the Cerenkov radiation produces a wave front with an apex half-angle of \(53.0^{\circ} .\) Calculate the speed of the electrons in the water. (The speed of light in water is \(2.25 \times 10^{8} \mathrm{m} / \mathrm{s} .\) )

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A firework charge is detonated many meters above the ground. At a distance of \(400 \mathrm{m}\) from the explosion, the acoustic pressure reaches a maximum of \(10.0 \mathrm{N} / \mathrm{m}^{2}\) Assume that the speed of sound is constant at \(343 \mathrm{m} / \mathrm{s}\) throughout the atmosphere over the region considered, that the ground absorbs all the sound falling on it, and that the air absorbs sound energy as described by the rate \(7.00 \mathrm{dB} / \mathrm{km} .\) What is the sound level (in \(\mathrm{dB}\) ) at \(4.00 \mathrm{km}\) from the explosion?

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