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Chapter 16: Problem 69

The bellow of a territorial bull hippopotamus has been measured at \(115 \mathrm{dB}\) above the threshold of hearing. What is the sound intensity?

Short Answer

Expert verified
The sound intensity is approximately 0.316 W/m虏.

Step by step solution

01

Understand the Decibel Scale

The decibel (dB) scale is a logarithmic scale used to measure sound intensity. The threshold of hearing is the quietest sound that the average human ear can hear, which is set at 0 dB. Each increase of 10 dB represents a tenfold increase in intensity.
02

Identify the Given Information

In this exercise, you are given a sound level of 115 dB above the threshold of hearing, which is set at 0 dB.
03

Use the Formula for Sound Intensity Level

The formula to convert a sound level in decibels to sound intensity is given by:\[L = 10 \cdot \log_{10}\left(\frac{I}{I_0}\right)\]where:- \(L\) is the sound level in decibels,- \(I\) is the intensity of the sound,- \(I_0\) is the reference intensity, set at the threshold of hearing, \(1 \times 10^{-12} \text{ W/m}^2\).
04

Plug in the Given Values

Given that \(L = 115\, \text{dB}\), the equation becomes:\[115 = 10 \cdot \log_{10}\left( \frac{I}{1 \times 10^{-12}} \right)\]
05

Solve for Sound Intensity \(I\)

Divide both sides by 10:\[11.5 = \log_{10}\left( \frac{I}{1 \times 10^{-12}} \right)\]Convert the logarithmic equation to an exponential equation:\[10^{11.5} = \frac{I}{1 \times 10^{-12}}\]Multiply both sides by \(1 \times 10^{-12}\) to solve for \(I\):\[I = 10^{11.5} \times 1 \times 10^{-12}\]Finally, calculate \(I\):\[I = 10^{-0.5} = 3.16 \times 10^{-1} \text{ W/m}^2\]
06

Conclusion

The sound intensity of the bull hippopotamus's bellow is approximately \(0.316 \text{ W/m}^2\).

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Key Concepts

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

Decibel Scale
The decibel scale is a fascinating yet simple way to measure sound intensity levels. At its core, it's a logarithmic scale, which means it doesn't increase in a linear fashion. Instead, each increment represents a tenfold increase in intensity.
The scale is widely used because it efficiently handles the vast range of intensities the human ear can perceive. Starting at 0 dB, which represents the threshold of hearing or the quietest sound an average human ear can pick up.
When we say a sound is at 115 dB, it means it is significantly louder than the threshold, specifically over a trillion times more intense.
  • Logarithmic nature simplifies comparison of vastly different intensities.
  • The threshold of hearing is the reference for this scale, set at 0 dB.
  • Each 10 dB step equals a tenfold increase in intensity.
Threshold of Hearing
The threshold of hearing is a key concept in understanding how we perceive sound. It is the quietest sound that the average human ear can detect and is set at 0 dB on the decibel scale. This reference point acts as a baseline for measuring other sound levels.
The threshold corresponds to a sound intensity of about \(1 \times 10^{-12} \text{ W/m}^2\), a very faint sound, such as the rustling of leaves.
In practical terms:
  • The threshold of hearing provides a starting point for the decibel scale.
  • All other sounds are measured against this baseline.
  • This helps to quantify different sound levels experienced in everyday life.
Exponential Equation
Exponential equations are integral in transforming logarithmic data, like decibels, back into standard intensity values. They enable us to convert a logarithmic expression into a multiplication operation.
In sound intensity measurement, the formula \[L = 10 \cdot \log_{10}\left(\frac{I}{I_0}\right)\]shows how decibels relate to intensity. By rearranging the equation, we use exponential functions to find actual sound intensities.
To solve the equation when the sound level is given, we:
  • Convert the logarithmic equation using the exponential function.
  • Set the base as 10, since we're dealing with a log to base 10.
  • Utilize the exponential equation to find that \(10^{11.5} \cdot 1 \times 10^{-12}\) equals the intensity \(I\).
  • This approach turns complex log expressions into relatable values, such as \(0.316 \text{ W/m}^2\).

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Most popular questions from this chapter

A wave traveling in the \(+x\) direction has an amplitude of \(0.35 \mathrm{m}\), a speed of \(5.2 \mathrm{m} / \mathrm{s},\) and a frequency of \(14 \mathrm{Hz}\). Write the equation of the wave in the form given by either Equation 16.3 or 16.4

A convertible moves toward you and then passes you; all the while, its loudspeakers are producing a sound. The speed of the car is a constant \(9.00 \mathrm{m} / \mathrm{s},\) and the speed of sound is \(343 \mathrm{m} / \mathrm{s} .\) What is the ratio of the frequency you hear while the car is approaching to the frequency you hear while the car is moving away?

A hunter is standing on flat ground between two vertical cliffs that are directly opposite one another. He is closer to one cliff than to the other. He fires a gun and, after a while, hears three echoes. The second echo arrives \(1.6 \mathrm{s}\) after the first, and the third echo arrives \(1.1 \mathrm{s}\) after the second. Assuming that the speed of sound is \(343 \mathrm{m} / \mathrm{s}\) and that there are no reflections of sound from the ground, find the distance between the cliffs.

A dish of lasagna is being heated in a microwave oven. The effective area of the lasagna that is exposed to the microwaves is \(2.2 \times\) \(10^{-2} \mathrm{m}^{2} .\) The mass of the lasagna is \(0.35 \mathrm{kg},\) and its specific heat capacity is \(3200 \mathrm{J} / \mathrm{kg} \cdot \mathrm{C}^{\circ}\) ). The temperature rises by \(72^{\circ} \mathrm{C}\) in 8.0 minutes. What is the intensity of the microwaves in the oven?

To measure the acceleration due to gravity on a distant planet, an astronaut hangs a \(0.055-\mathrm{kg}\) ball from the end of a wire. The wire has a length of \(0.95 \mathrm{m}\) and a linear density of \(1.2 \times 10^{-1} \mathrm{kg} / \mathrm{m} .\) Using electronic equipment, the astronaut measures the time for a transverse pulse to travel the length of the wire and obtains a value of 0.016 s. The mass of the wire is negligible compared to the mass of the ball. Determine the acceleration due to gravity.
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