/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 48 One cue your hearing system uses... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 20: Problem 48

One cue your hearing system uses to localize a sound ( \(1 . e,\) to tell where a sound is coming from) is the slight difference in the arrival times of the sound at your ears. Your ears are spaced approximately \(20 \mathrm{cm}\) apart. Consider a sound source \(5.0 \mathrm{m}\) from the center of your head along a line \(45^{\circ}\) to your right. What is the difference in arrival times? Give your answer in microseconds. Hint: You are looking for the difference between two numbers that are nearly the same. What does this near equality imply about the necessary precision during intermediate stages of the calculation?

Short Answer

Expert verified
The difference in arrival times is 20 microseconds.

Step by step solution

01

Calculate the distance to each ear

Start by calculating the distance (d1) to the right ear and the distance (d2) to the left ear. The distance to the right ear is simply the radius of the circle, which is 5.0 m. To find the distance to the left ear, use Pythagoras' theorem since the problem can be viewed as a right-angled triangle. The hypotenuse is the distance to the left ear, the opposite side to the angle of 45掳 is the distance to the center of the head to the sound source (5.0m), and the adjacent side to the angle of 45掳 is the half of the distance between the two ears (10cm). So \(d2 = \sqrt{(d/2^2 + r^2)} = \sqrt{(0.10m^2 + 5.0m^2)} = 5.005m\).
02

Calculate the time it takes for the sound to reach each ear

Use the formula for velocity to calculate the time it takes for the sound to reach each ear. The speed of sound in dry air at 20 掳C (293.15 K) is 343.2 meters per second. The time it takes for the sound to reach the right ear (t1) is equal to the distance to the right ear (d1) divided by the speed of sound (v). So \(t1 = d1/v = 5.0m / 343.2m/s = 0.01457s = 14.57ms\). The same goes for the time it takes for the sound to reach the left ear (t2) using the distance to the left ear (d2). So \(t2 = d2/v = 5.005m / 343.2m/s = 0.01459s = 14.59ms\).
03

Calculate the difference in arrival times

Finally, subtract the time it takes for the sound to reach the right ear (t1) from the time it takes for the sound to reach the left ear (t2) to get the difference in arrival times. So \(\Delta t = t2 - t1 = 14.59ms - 14.57ms = 0.02ms = 20 \mu s\) .

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Interaural Time Difference
Understanding how we pinpoint a sound's origin involves the concept of the interaural time difference (ITD). Essentially, this is the slight time difference of sound reaching each ear. When a sound is closer to one ear, it arrives there first, helping our brains determine the direction of the sound. For humans, even milliseconds matter in perceiving and localizing sounds accurately. In the example above, sound reaches the right ear slightly faster than the left, creating an ITD of roughly 20 microseconds.
Physics Problem Solving
Solving physics problems like this one involves breaking the problem into manageable steps. In the given exercise, it's crucial to first understand the geometrical setup. Whether it's creating diagrams or thoroughly reading the problem, understanding your goal is essential. After getting the concept, follow these steps:
  • Determine all known values, such as distances and degrees.
  • Use relevant formulas鈥攈ere, the velocity of sound and distances.
  • Perform calculations with care, especially when determining differences that are minuscule.
Applying this structured approach aids in tackling not only sound localization queries but many physics challenges.
Acoustics
The science of acoustics covers how sound travels through various media, the factors influencing this travel, and how we perceive it. Sound travels as waves and in air, these waves move at standard speeds dependent on the air temperature and other factors. At 20掳C or 293.15 K, sound moves at approximately 343.2 meters per second. This consistency in speed allows calculations and predictions, like finding how sound arrives at each ear. In scenarios where precision is key, such as determining interaural delays, accurate speed values are critical.
Pythagorean Theorem
The Pythagorean theorem is a cornerstone in connecting geometry with real-world computations. It provides a way to calculate one side of a triangle when the other two sides are known, especially applicable in right-angle triangles. In the context of the exercise, the distance calculation from the sound source to the left ear uses this theorem to handle the right triangle with a 45掳 angle. The formula, \(a^2 + b^2 = c^2\), where \(c\) is the hypotenuse, helps us derive that \(d2 = \sqrt{(0.10^2 + 5.0^2)}\), contributing to precise auditory measurements. Such applications of geometric principles are vital in both physics and other scientific domains.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The sound intensity level \(5.0 \mathrm{m}\) from a large power saw is 100 dB. At what distance will the sound be a more tolerable 80 \(\mathrm{dB} ?\)

Write the displacement equation for a sinusoidal wave that is traveling in the negative \(y\) -direction with wavelength \(50 \mathrm{cm}\) speed \(4.0 \mathrm{m} / \mathrm{s},\) and amplitude \(5.0 \mathrm{cm} .\) Assume \(\phi_{0}=0\).

A sound wave is described by \(D(y, t)=(0.0200 \mathrm{mm}) \times\) \(\sin [(8.96 \mathrm{rad} / \mathrm{m}) \mathrm{y}+(3140 \mathrm{rad} / \mathrm{s}) t+\pi / 4 \mathrm{rad}],\) where \(\mathrm{y}\) is in \(\mathrm{m}\) and \(t\) is in \(s\) a. In what direction is this wave traveling? b. Along which axis is the air oscillating? c. What are the wavelength, the wave speed, and the period of oscillation? d. Draw a displacement-versus-time graph \(D(y=1.00 \mathrm{m}, t)\) at \(y=1.00 \mathrm{m}\) from \(t=0 \mathrm{s}\) to \(t=4.00 \mathrm{ms}\).

A starship approaches its home planet at a speed of \(0.1 c .\) When it is \(54 \times 10^{6} \mathrm{km}\) away, it uses its green laser beam \((\lambda=540 \mathrm{nm})\) to signal its approach. a. How long does the signal take to travel to the home planet? b. At what wavelength is the signal detected on the home planet?

A sound wave with intensity \(2.0 \times 10^{-3} \mathrm{W} / \mathrm{m}^{2}\) is perceived to be modestly loud. Your eardrum is 6.0 mm in diameter. How much energy will be transferred to your eardrum while listening to this sound for \(1.0 \mathrm{min} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Solid State Physics

Read Explanation

Turning Points in Physics

Read Explanation

Medical Physics

Read Explanation

Astrophysics

Read Explanation

Quantum Physics

Read Explanation

Work Energy and Power

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.