/*! 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 1 BID (a) Music. When a person sin... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

BID (a) Music. When a person sings, his or her vocal cords vibrate in a repetitive pattern that has the same frequency as the note that is sung. If someone sings the note \(B\) flat, which has a fre- quency of 466 Hz, how much time does it take the person's vocal cords to vibrate through one complete cycle, and what is the angu- lar frequency of the cords? (b) Hearing. When sound waves strike the eardrum, this membrane vibrates with the same frequency as the sound. The highest pitch that typical humans can hear has a period of 50.0\(\mu \mathrm{s} .\) What are the frequency and angular frequency of the vibrating eardrum for this sound? (c) Vision. When light having vibrations with angular frequency ranging from \(2.7 \times 10^{15} \mathrm{rad} / \mathrm{s}\) to \(4.7 \times 10^{15} \mathrm{rad} / \mathrm{s}\) strikes the retina of the eye, it stimulates the receptor cells there and is perceived as visible light. What are the limits of the period and frequency of this light? (d) Ultrasound. High-frequency sound waves (ultrasound) are used to probe the interior of the body, much as x rays do. To detect small objects such as tumors, a frequency of around 5.0 \(\mathrm{MHz}\) is used. What are the period and angular frequency of the molecular vibrations caused by this pulse of sound?

Short Answer

Expert verified
(a) 0.00215 s, 2927 rad/s; (b) 20 kHz, 125664 rad/s; (c) 2.33-1.34 x 10^{-16} s, 4.30-7.48 x 10^{14} Hz; (d) 2.0 x 10^{-7} s, 3.14 x 10^7 rad/s.

Step by step solution

01

Calculate the time for one cycle for note B flat

The frequency of the note B flat is given as 466 Hz. The time period (T) is the reciprocal of the frequency (f), so we use the formula:\[T = \frac{1}{f} = \frac{1}{466} \text{ seconds} \approx 0.00215 \text{ seconds}\].
02

Compute the angular frequency for note B flat

Angular frequency (ω) is given by the formula:\[\omega = 2 \pi f = 2 \pi \times 466 \text{ rad/s} \approx 2927 \text{ rad/s}\].
03

Determine the frequency for the highest audible pitch

The period of the highest pitch is given as 50.0 \(\mu\text{s}\) or \(50.0 \times 10^{-6}\text{ s}\). To find the frequency (f), use the formula:\[f = \frac{1}{T} = \frac{1}{50.0 \times 10^{-6}} \text{ Hz} = 20,000 \text{ Hz} or 20 \text{ kHz}\].
04

Calculate the angular frequency for the highest pitch

Using the frequency calculated in the previous step, determine the angular frequency (ω):\[\omega = 2 \pi f = 2 \pi \times 20000 \text{ rad/s} \approx 125664 \text{ rad/s}\].
05

Find the period limits for visible light

The angular frequency limits for visible light are given. For the lower limit of \(2.7 \times 10^{15} \text{ rad/s}\):\[T = \frac{2\pi}{\omega} = \frac{2\pi}{2.7 \times 10^{15}} \text{ s} \approx 2.33 \times 10^{-16} \text{ s}\],and for the upper limit of \(4.7 \times 10^{15} \text{ rad/s}\):\[T = \frac{2\pi}{4.7 \times 10^{15}} \text{ s} \approx 1.34 \times 10^{-16} \text{ s}\].
06

Compute the frequency limits for visible light

For the lower angular frequency limit of \(2.7 \times 10^{15} \text{ rad/s}\):\[f = \frac{\omega}{2\pi} = \frac{2.7 \times 10^{15}}{2\pi} \text{ Hz} \approx 4.30 \times 10^{14} \text{ Hz}\],and for the upper limit \(4.7 \times 10^{15} \text{ rad/s}\):\[f = \frac{4.7 \times 10^{15}}{2\pi} \text{ Hz} \approx 7.48 \times 10^{14} \text{ Hz}\].
07

Calculate the period for ultrasound's high frequency

The frequency of the ultrasound is given as 5.0 MHz or \(5.0 \times 10^6 \text{ Hz}\). Use the reciprocal to find the period (T):\[T = \frac{1}{f} = \frac{1}{5.0 \times 10^6} \text{ s} \approx 2.0 \times 10^{-7} \text{ s}\].
08

Determine the angular frequency for ultrasound

With the given frequency, find the angular frequency (ω):\[\omega = 2 \pi f = 2 \pi \times 5.0 \times 10^6 \text{ rad/s} \approx 3.14 \times 10^7 \text{ rad/s}\].

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 91Ó°ÊÓ!

Key Concepts

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

Frequency: The Rhythm of Oscillations
Frequency is a fundamental concept in the study of oscillations, describing how often a repeating event occurs per unit of time. It is measured in Hertz (Hz), where 1 Hz equals one cycle per second.

Key points to remember:
  • Frequency is the inverse of the time period (T). The relationship is given by the formula: \[f = \frac{1}{T}\]
  • In this context, the frequency of a musical note or sound wave indicates how many times the wave pattern repeats each second.
  • If someone sings a note with a frequency of 466 Hz, their vocal cords complete 466 cycles per second.
This same principle applies to the frequency of sound waves hitting your eardrum, which mimics the frequency of the sound heard. For example, a sound with a frequency of 20 kHz (20,000 Hz) hits the eardrum 20,000 times per second, indicating a high-pitched sound.
Understanding frequency helps us dive into a wide range of phenomena, from hearing musical notes to perceiving complex sound from our environment.
Angular Frequency: Going Around in Circles
Angular frequency, often denoted by the Greek letter \(\omega\), is another vital concept in the study of oscillations. It's particularly associated with rotating objects or oscillating systems and measures how fast something moves through a circular path. Angular frequency is expressed in radians per second (rad/s).

Let's explore its relationships:
  • Angular frequency is linked to frequency through the formula \[\omega = 2\pi f\]
    This shows the conversion between linear frequency in regular cycles to angular movement in radians.
  • When analyzing sounds, the angular frequency tells us how rapidly the waveform rotates through each cycle.
For instance, a note B flat at 466 Hz which, when converted, has an angular frequency about 2927 rad/s.
Angular frequency provides crucial insights into circular motion and vibrations, paving a path to understanding more complex oscillatory behavior.
Vibration: The Dynamic Pulse of Motion
Vibrations occur when something moves back and forth rapidly. In the physical world, vibrations can cause elements like vocal cords or eardrums to constantly fluctuate around a certain position, producing sound and other phenomena.

Consider these aspects of vibration:
  • Vibrational motion is characterized by aspects like frequency and angular frequency, which dictate how fast and in what manner the object moves back and forth.
  • In practical applications, vibrations range from the sinusoidal movements of sound waves to the complex patterns of light waves that strike the retina.
The vibrations of vocal cords when you sing, for instance, translate into sound waves perceived as musical notes.

Exploring vibrations is essential in diverse fields, from music and engineering to understanding biological sensory processes like hearing and sight.

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

CP On the planet Newtonia, a simple pendulum having a bob with mass 1.25 \(\mathrm{kg}\) and a length of 185.0 \(\mathrm{cm}\) takes 1.42 \(\mathrm{s}\) . when released from rest, to swing through an angle of \(12.5^{\circ}\) , where it again has zero speed. The circumference of Newtonia is measured to be \(51,400 \mathrm{km}\) . What is the mass of the planet Newtonia?

CP A rifle bullet with mass 8.00 \(\mathrm{g}\) and initial horizontal velocity 280 \(\mathrm{m} / \mathrm{s}\) strikes and embeds itself in a block with mass 0.992 \(\mathrm{kg}\) that rests on a frictionless surface and is attached to one end of an ideal spring. The other end of the spring is attached to the wall. The impact compresses the spring a maximum distance of 18.0 \(\mathrm{cm}\) . After the impact, the block moves in SHM. Calculate the period of this motion.

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the amplitude of the motion is \(0.090 \mathrm{m},\) it takes the block 2.70 s to travel from \(x=0.090 \mathrm{m}\) to \(x=-0.090 \mathrm{m} .\) If the amplitude is doubled, to \(0.180 \mathrm{m},\) how long does it take the block to travel (a) from \(x=0.180 \mathrm{m}\) to \(x=-0.180 \mathrm{m}\) and (b) from \(x=0.090 \mathrm{m}\) to \(x=-0.090 \mathrm{m} ?\)

We want to hang a thin hoop on a horizontal nail and have the hoop make one complete small-angle oscillation each 2.0 \(\mathrm{s}\) What must the hoop's radius be?

Inside a NASA test vehicle, a \(3.50-\) kg ball is pulled along by a horizontal ideal spring fixed to a friction-free table. The force constant of the spring is 225 \(\mathrm{N} / \mathrm{m}\) . The vehicle has a steady acceleration of \(5.00 \mathrm{m} / \mathrm{s}^{2},\) and the ball is not oscillating. Suddenly, when the vehicle's speed has reached \(45.0 \mathrm{m} / \mathrm{s},\) its engines turn off, thus eliminating its acceleration but not its velocity. Find (a) the amplitude and (b) the frequency of the resulting oscillations of the ball. (c) What will be the ball's maximum speed relative to the vehicle?

See all solutions

Recommended explanations on Physics Textbooks

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.