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(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 frequency of 466 Hz, how much time does it take the person's vocal cords to vibrate through one complete cycle, and what is the angular 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 young humans can hear has a period of 50.0 μs. 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 × 1015 rad/s to 4.7 × 1015 rad/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 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 per cycle, 2928.3 rad/s; (b) 20,000 Hz, 125,664 rad/s; (c) 2.34 to 1.34 fs, 430 to 748 THz; (d) 0.2 µs, 31.42 rad/µs.

Step by step solution

01

Time for One Cycle of Vocal Cord Vibration

For a note with a frequency of 466 Hz, we find the time for one complete vibration cycle using the formula T=1f, where T is the period and f is the frequency in hertz. For B flat, f=466 Hz. So, T=14660.00215 s per cycle.
02

Angular Frequency of Vocal Cord Vibration

Angular frequency ω is calculated using ω=2πf. Given f=466 Hz, ω=2π×4662928.3 rad/s.
03

Frequency of Eardrum Vibration

The frequency f can be found using f=1T. Here, the period T=50.0 μs = 50.0×106 s. So, f=150.0×106=20,000 Hz.
04

Angular Frequency of Eardrum Vibration

The angular frequency is calculated using ω=2πf. With f=20,000 Hz, ω=2π×20,000125,664 rad/s.
05

Limits of Period of Visible Light

The period T is the inverse of frequency. We calculate the limits using ω=2πf. For the lower limit ω=2.7×1015, f=2.7×10152π4.3×1014 Hz, so T=14.3×10142.34×1015 s. Similarly, for ω=4.7×1015, f7.48×1014 Hz, T1.34×1015 s.
06

Limits of Frequency of Visible Light

The frequency limits we calculated are approximately 4.3×1014 Hz and 7.48×1014 Hz based on the given angular frequency limits.
07

Period of Ultrasound Vibration

For a frequency of 5.0 MHz, the period T is 15.0×106=2.0×107 s.
08

Angular Frequency of Ultrasound Vibration

To find the angular frequency, use ω=2πf. With f=5.0×106 Hz, we get ω=2π×5.0×10631.42×106 rad/s.

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Key Concepts

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

Angular Frequency
Angular frequency is a key concept when discussing wave phenomena. It refers to how fast something oscillates or cycles in terms of radians per second. This measure is particularly useful in physics because it gives us a way to describe circular motion and wave frequencies in a more natural way for mathematical manipulations. To calculate angular frequency, use the formula ω=2πf, where ω represents the angular frequency and f is the frequency in hertz.

This relationship helps us understand phenomena like how sound waves, light waves, and vibrations interact with different media. Angular frequency is often used to connect linear frequency to rotational or wave phenomena, illustrating how many radians a vibrating system completes per second. This makes it fundamentally different but closely related to the regular frequency, which counts cycles per second.
Sound Waves
Sound waves are mechanical waves that travel through a medium such as air, water, or solids. These waves are created by vibrating objects, like vocal cords in humans, and consist of compressions and rarefactions that move through the air. A sound wave's frequency determines its pitch, which is why sounds can range from the low rumbling of a voice to high-pitched chirps.

The human ear is adapted to perceive a wide range of sound wave frequencies. Young humans can typically hear from around 20 Hz up to 20,000 Hz. This hearing range explains why some sounds are inaudible to adults but can be perceived by children or animals. Since the speed of sound in air is around 343 meters per second, the wavelength and frequency of a sound wave determine how sound behaves as it travels.
  • Low frequencies produce deeper sounds (bass).
  • High frequencies produce sharper, higher-pitched sounds (treble).
The interplay between frequency and period (the time it takes for one complete cycle) is crucial for understanding sound properties. The eardrum vibrates with the same frequency as the incoming sound wave, allowing us to enjoy music and communicate through speech.
Visible Light Spectrum
The visible light spectrum is the part of the electromagnetic spectrum that is visible to the human eye. It encompasses light waves with angular frequencies from approximately 2.7×1015 to 4.7×1015 radians per second. These light waves have a frequency range that correlates with colors perceived from violet to red. The variation in frequencies across this spectrum gives rise to different colors:
  • Higher frequencies correspond to the violet and blue side.
  • Lower frequencies correspond to the red side.
When light strikes the retina, its frequency and angular frequency send signals to the brain, which is perceived as color. Understanding these limits is crucial for applications in optics and vision sciences.

Unlike sound waves, light waves do not require a medium to travel, which is why we can see sunlight from the space vacuum. Also noteworthy is the period of these waves. The period is incredibly tiny, which indicates the rapid oscillation of light waves as they move, explaining why light travels so quickly.
Ultrasound
Ultrasound refers to sound waves with frequencies above the audible range of human hearing, typically above 20,000 Hz. In medical diagnostics, ultrasound waves use frequencies around 5 MHz to look inside the body, similar to how x-rays work but without the exposure to radiation.

Experts calculate the period of these waves to be very short, often in microseconds, due to their high frequency. This property makes ultrasound excellent for creating detailed images of internal organs and tissues because high-frequency sound waves produce better resolution images. The angular frequency of ultrasound can be calculated using the formula ω=2πf, which helps technicians understand the ultrasound's impact on what's being examined.
  • Medical professionals use ultrasound to detect changes in tissues.
  • It helps in procedures like prenatal scanning.
  • It's a non-invasive and safe way to explore internal conditions.
The use of such high frequencies allows ultrasound to detect even minute details, like tiny tumors, hence its preference in diagnostic imaging.

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Most popular questions from this chapter

A 0.0200-kg bolt moves with SHM that has an amplitude of 0.240 m and a period of 1.500 s. The displacement of the bolt is +0.240 m when t= 0. Compute (a) the displacement of the bolt when t= 0.500 s; (b) the magnitude and direction of the force acting on the bolt when t= 0.500 s; (c) the minimum time required for the bolt to move from its initial position to the point where x=0.180 m; (d) the speed of the bolt when x=0.180 m.

A 50.0-g hard-boiled egg moves on the end of a spring with force constant k= 25.0 N/m. Its initial displacement is 0.300 m. A damping force Fx=bvx acts on the egg, and the amplitude of the motion decreases to 0.100 m in 5.00 s. Calculate the magnitude of the damping constant b.

A block with mass m= 0.300 kg is attached to one end of an ideal spring and moves on a horizontal frictionless surface. The other end of the spring is attached to a wall. When the block is at x=+0.240 m, its acceleration is ax=12.0 m/s2 and its velocity is vx=+4.00 m/s. What are (a) the spring's force constant k; (b) the amplitude of the motion; (c) the maximum speed of the block during its motion; and (d) the maximum magnitude of the block's acceleration during its motion?

The balance wheel of a watch vibrates with an angular amplitude Θ, angular frequency ω, and phase angle ϕ= 0. (a) Find expressions for the angular velocity dθ/dt and angular acceleration d2θ/dt2 as functions of time. (b) Find the balance wheel's angular velocity and angular acceleration when its angular displacement is Θ, and when its angular displacement is Θ/2 and θ is decreasing. (Hint: Sketch a graph of θ versus t.)

When a body of unknown mass is attached to an ideal spring with force constant 120 N/m, it is found to vibrate with a frequency of 6.00 Hz. Find (a) the period of the motion; (b) the angular frequency; (c) the mass of the body.

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