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Chapter 14: Problem 55

If you double your distance from a point source of sound, by what factor does the intensity change? Explain.

Short Answer

Expert verified
The intensity becomes one-quarter when the distance is doubled.

Step by step solution

01

Understanding Intensity and Distance

Sound intensity is inversely proportional to the square of the distance from the source. This means that the farther you are from the sound source, the lower the intensity you will experience.
02

Applying the Inverse Square Law

The inverse square law for sound intensity can be expressed as \( I \propto \frac{1}{r^2} \), where \( I \) is the intensity and \( r \) is the distance from the source. Doubling the distance will affect intensity.
03

Calculating New Intensity

If the distance \( r \) is doubled (i.e., \( r ightarrow 2r \)), substitute this into the inverse square law: \( I ightarrow \frac{1}{(2r)^2} = \frac{1}{4r^2} \). Thus, the new intensity \( I' = \frac{I}{4}\).
04

Conclusion

By doubling the distance from the sound source, the intensity of the sound reduces to one-quarter of its original intensity due to the inverse square relationship.

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Key Concepts

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

Sound Intensity
Sound intensity is a measure of the power of sound per unit area. It's one of the ways to describe how strong a sound is. Remember when you've been near a loudspeaker at a concert? That's high sound intensity. Now think about how the music feels quieter as you move away. That's sound intensity decreasing. The unit of sound intensity is typically watts per square meter (W/m虏).
Sound intensity is crucial in physics because it describes how energy in a sound wave spreads out from the source.
  • It helps in understanding how sound diminishes over distance.
  • It lets us quantify sound levels in various environments.
Understanding sound intensity is foundational for diving deeper into how sound behaves over distances and through different media.
Distance and Intensity Relationship
The relationship between distance and sound intensity is a fascinating aspect of physics. It's governed by the inverse square law, which says that intensity decreases with the square of the distance from the source.
Imagine holding a flashlight. The closer you are to a wall, the brighter and more concentrated the light appears. Move further away, and the light spreads out, getting dimmer. This is similar to how sound works:
  • The intensity halves not when you double your distance, but actually reduces to one-fourth (or a quarter).
  • This happens because the sound wavefronts spread over a larger area as they move outward.
When you're twice as far away, you're sharing the same sound energy over four times the area, thus the intensity drops to a quarter. This natural spread of sound is why whispers get lost in a large room!
Physics Problem Solving
Solving physics problems like understanding the inverse square law in sound requires a strategic approach. Here's how you can tackle such problems efficiently:
  • **Identify the Variables:** Start by distinguishing between what's given and what's needed. For instance, know that distance and intensity are key.
  • **Understand the Relationships:** Grasp the foundational laws, like how doubling distance affects intensity in this case.
  • **Apply Mathematical Principles:** Use relevant formulas, such as the inverse square law expressed as \( I \propto \frac{1}{r^2} \).
  • **Step-by-Step Calculation:** Break down the process. If you're doubling the distance, calculate how the formula predicts intensity reduces to a quarter, as with \( I' = \frac{I}{4} \).
  • **Conclude Clearly:** Summarize your findings. Confirm that your calculated outcomes align with expected physical behavior.
Engaging with physics problems in this methodical way encourages deeper understanding and helps solidify principles in your mind, paving the road to mastering the subject!

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Most popular questions from this chapter

A northern mockingbird sings a single note with a frequency \(220 \mathrm{~Hz}\) as it flies directly toward you. Is the frequency you hear greater than, less than, or equal to \(220 \mathrm{~Hz}\) ? Explain.

What is the wavelength of the second harmonic in a \(2.5\)-m-long pipe that is open at both ends?

The human ear canal is much like an organ pipe that is closed at one end (at the tympanic membrane, or eardrum) and open at the other (see Figure 14.20). A typical ear canal has a length of about \(2.4 \mathrm{~cm}\). (a) What are the fundamental frequency of the ear canal and the wavelength of that standing wave? (b) Find the frequency and wavelength of the ear canal's third harmonic. (Recall that the third harmonic in this case is the standing wave with the second-lowest frequency.) (c) Suppose a person has an ear canal that is shorter than \(2.4 \mathrm{~cm}\). Is the fundamental frequency of that person's ear canal greater than, less than, or the same as the value found in part (a)? Explain. [Notice that the frequencies found in parts (a) and (b) correspond closely to the frequencies of enhanced sensitivity in Figure 14.19.]

Dolphins of the open ocean are classified as Type II Odontocetes (toothed whales). These animals use ultrasonic "clicks" with a frequency of \(55 \mathrm{kHz}\) to navigate and find prey. (a) Suppose a dolphin sends out a series of clicks that are reflected back from the bottom of the ocean \(75 \mathrm{~m}\) below. How much time elapses before the dolphin hears the echoes of the clicks? (The speed of sound in seawater is approximately \(1530 \mathrm{~m} / \mathrm{s}\) ) (b) What is the wavelength of a \(55-\mathrm{kHz}\) sound in the ocean?

An emergency vehicle blowing its siren is moving away from you with a speed of \(23 \mathrm{~m} / \mathrm{s}\). The sound you hear has a frequency of \(590 \mathrm{~Hz}\). What is the frequency produced by the siren?
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