Problem 1 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

(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 1: 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:\[T = \frac{1}{f} = \frac{1}{466} \text{ seconds} \approx 0.00215 \text{ seconds}\].

Step 2: Compute the angular frequency for note B flat

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

Step 3: 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:\[f = \frac{1}{T} = \frac{1}{50.0 \times 10^{-6}} \text{ Hz} = 20,000 \text{ Hz} or 20 \text{ kHz}\].

Step 4: Calculate the angular frequency for the highest pitch

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

Step 5: 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 \(4.7 \times 10^{15} \text{ rad/s}\):\[T = \frac{2\pi}{4.7 \times 10^{15}} \text{ s} \approx 1.34 \times 10^{-16} \text{ s}\].

Step 6: 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 \(4.7 \times 10^{15} \text{ rad/s}\):\[f = \frac{4.7 \times 10^{15}}{2\pi} \text{ Hz} \approx 7.48 \times 10^{14} \text{ Hz}\].

Step 7: 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):\[T = \frac{1}{f} = \frac{1}{5.0 \times 10^6} \text{ s} \approx 2.0 \times 10^{-7} \text{ s}\].

Step 8: Determine the angular frequency for ultrasound

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

Key Concepts

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.