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A baby's mouth is 30 cm from her father's ear and 1.50 m from her mother's ear. What is the difference between the sound intensity levels heard by the father and by the mother?

Short Answer

Expert verified
The difference in sound intensity levels is approximately 13.98 dB.

Step by step solution

01

Understand the Problem

We need to calculate the difference in sound intensity levels between two distances from a sound source: 30 cm (0.30 m) and 1.50 m. The sound intensity level is measured in decibels (dB). The formula for sound intensity level is related to the distance from the sound source.
02

Use the Inverse Square Law

The sound intensity is inversely proportional to the square of the distance from the source. This means the sound intensity at a distance \( r \) is given by \( I \propto \frac{1}{r^2} \). To compare two distances, we need to calculate the ratio of their intensities.
03

Calculate Intensity Ratio

Let \( I_1 \) be the intensity at 0.30 m and \( I_2 \) be the intensity at 1.50 m. The ratio \( \frac{I_1}{I_2} \) is equal to \( \left( \frac{1.50}{0.30} \right)^2 = 25 \). This means the intensity is 25 times stronger at 0.30 m than at 1.50 m.
04

Calculate Difference in Sound Intensity Levels

The difference in intensity levels \( \Delta L \) in decibels is calculated using the formula \( \Delta L = 10 \log_{10} \left( \frac{I_1}{I_2} \right) \). Plugging in the values, we get \( \Delta L = 10 \log_{10}(25) \approx 10 \times 1.398 = 13.98 \text{ dB} \).
05

Provide the Result

The difference in the sound intensity levels heard by the father and the mother is approximately 13.98 dB.

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Key Concepts

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

Inverse Square Law
In physics, the inverse square law describes how a particular physical quantity or strength decreases with increasing distance from the source. It's a crucial concept in understanding how sound intensity changes with distance.
  • When applied to sound, intensity is inversely proportional to the square of the distance from the source.
  • This means that as you move away from the sound source, the intensity decreases rapidly, not in a linear fashion but rather quadratically.
For instance, in our problem, as the father is closer to the baby's mouth than the mother, he perceives a higher sound intensity due to this law.
Let's break this down mathematically. If the sound intensity at a distance \( r \) is represented as \( I \), then:\[ I \propto \frac{1}{r^2} \]This means if the distance is doubled, the intensity becomes one-fourth its original value. Thus, small variations in distance can significantly affect the sound listeners perceive.
Decibels
Decibels (dB) are a logarithmic unit used to measure the intensity of sound. It's a more practical way to express vast differences in sound intensity, which can range over several orders of magnitude.
  • Using a logarithmic scale helps compress the large range of sound intensities into a manageable scale.
  • This measurement is relative and often compares two intensity levels.
The formula used to convert the intensity ratio to decibels is:\[ L = 10 \log_{10} \(\frac{I_1}{I_2}\) \]Here, \( L \) is the sound intensity level measured in decibels, and \( I_1 \) and \( I_2 \) are two intensity levels being compared.
In the exercise, we determined that the sound intensity easier perceived by the father is much higher compared to the mother, strengthening the baby's voice noticeably due to the exponential nature of decibels.
Sound Intensity Level
Sound intensity level gives a measure of the power of sound per unit area, where area is perpendicular to the direction of sound wave travel.
  • It’s important because it provides an effective way of comparing how sounds of differing intensities are perceived by the human ear.
  • The concept combines both intensity and listener perception, looking at how intense sounds are at the receiver's location.
In practical scenarios like our textbook problem, changes in distance directly affect the intensity level because of the inverse square law.
When comparing two sound intensity levels, we use the decibel scale to calculate the difference as seen:\[\Delta L = 10 \log_{10} \(\frac{I_1}{I_2}\) \]This formula tells us how different two sound levels are when perceived by listeners at varying distances. For the father and mother, this equation shows approximately a 13.98 dB difference, showcasing just how variably intense the sound can seem from two different perspectives.

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Most popular questions from this chapter

Horseshoe bats (genus \(Rhinolophus\)) emit sounds from their nostrils and then listen to the frequency of the sound reflected from their prey to determine the prey's speed. (The "horseshoe" that gives the bat its name is a depression around the nostrils that acts like a focusing mirror, so that the bat emits sound in a narrow beam like a flashlight.) A \(Rhinolophus\) flying at speed \(v_{bat}\) emits sound of frequency \(f_{bat}\); the sound it hears reflected from an insect flying toward it has a higher frequency \(f_{refl}\). (a) Show that the speed of the insect is $$vinsect = v\Bigg[\frac{f_{refl}(v - v_{bat}) - f_{bat}(v + v_{bat})}{f_{refl}(v - v_{bat}) + f_{bat}(v + v_{bat})}\Bigg] $$ where \(v\) is the speed of sound. (b) If \(f_{bat} =\) 80.7 kHz, \(f_{refl} =\) 83.5 kHz, and \(v_{bat} =\) 3.9 m/s, calculate the speed of the insect.

You live on a busy street, but as a music lover, you want to reduce the traffic noise. (a) If you install special soundreflecting windows that reduce the sound intensity level (in dB) by 30 dB, by what fraction have you lowered the sound intensity (in W\(/\)m\(^2\))? (b) If, instead, you reduce the intensity by half, what change (in dB) do you make in the sound intensity level?

The shock-wave cone created by a space shuttle at one instant during its reentry into the atmosphere makes an angle of 58.0\(^\circ\) with its direction of motion. The speed of sound at this altitude is 331 m/s. (a) What is the Mach number of the shuttle at this instant, and (b) how fast (in m/s and in mi/h) is it traveling relative to the atmosphere? (c) What would be its Mach number and the angle of its shock-wave cone if it flew at the same speed but at low altitude where the speed of sound is 344 m/s ?

A person is playing a small flute 10.75 cm long, open at one end and closed at the other, near a taut string having a fundamental frequency of 600.0 Hz. If the speed of sound is 344.0 m/s, for which harmonics of the flute will the string resonate? In each case, which harmonic of the string is in resonance?

You are trying to overhear a juicy conversation, but from your distance of 15.0 m, it sounds like only an average whisper of 20.0 dB. How close should you move to the chatterboxes for the sound level to be 60.0 dB?

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