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A piano string sounds a middle A by vibrating primarily at 220 \(\mathrm{Hz}\) , (a) Calculate the string's period and angular frequency. (b) Calculate the period and angular frequency for a soprano singing an A one octave higher, which is twice the frequency of the piano string.

Short Answer

Expert verified
The piano string's period is 0.00455 s and angular frequency is 1382.3 rad/s; the soprano's period is 0.00227 s and angular frequency is 2764.6 rad/s.

Step by step solution

01

Define the Concept of Period and Frequency

The period of a wave or vibration is the time it takes to complete one cycle, and it's the reciprocal of the frequency \( f \). Given the frequency of the piano string \( f = 220 \, \text{Hz} \), the period \( T \) can be calculated using the formula \( T = \frac{1}{f} \).
02

Calculate Piano String's Period

For the piano string with a frequency of 220 Hz, the period \( T \) is calculated as follows: \( T = \frac{1}{220} \approx 0.00455 \, \text{seconds} \).
03

Define Angular Frequency

Angular frequency \( \omega \) is given by the formula \( \omega = 2\pi f \), where \( f \) is the frequency of the vibration. This represents how many radians per second the wave oscillates.
04

Calculate Piano String's Angular Frequency

Using the formula for angular frequency \( \omega = 2\pi \times 220 \), the angular frequency is \[ \omega \approx 1382.3 \, \text{rad/s} \].
05

Calculate Frequency for Soprano's A

An octave higher frequency is twice the frequency of the original, so for the soprano's A note, the frequency is \( 2 \times 220 = 440 \, \text{Hz} \).
06

Calculate Soprano's A Period

The period for the soprano's A note is \( T = \frac{1}{440} \approx 0.00227 \, \text{seconds} \).
07

Calculate Soprano's A Angular Frequency

Using the formula for angular frequency with \( f = 440 \, \text{Hz} \), we have \( \omega = 2\pi \times 440 \), thus \[ \omega \approx 2764.6 \, \text{rad/s} \].

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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 for understanding the rate at which waves or oscillations repeat over time. In other words, it tells us how fast something is vibrating in terms of radians per second. An everyday example of this would be the vibration of a piano string. Just like frequency, angular frequency relates directly to the speed of the oscillation, but with a particular focus on the angular component, expressed in radians.

To calculate angular frequency, you use the formula \( \omega = 2 \pi f \), where \( f \) is the frequency of the wave. For instance, if a piano string has a frequency of 220 Hz, then its angular frequency can be determined by \( \omega = 2 \pi \times 220 \), which results in \( \omega \approx 1382.3 \text{ rad/s} \).
  • The formula incorporates \( 2 \pi \) because a complete cycle of a sine wave encompasses \( 2 \pi \) radians.
  • Angular frequency can often help to simplify the analysis of harmonic motion and is crucial in physics and engineering realms.
Wave Oscillation
Wave oscillation refers to the repetitive motion that waves undergo. This concept is crucial for understanding how waves behave over time. Think of waves on a string, like a guitar or piano string, that move up and down or back and forth repetitively.

Each complete movement of the wave from one starting point back to the same point is known as a cycle. The duration of this cycle is called the period, denoted by \( T \). Period and frequency are closely related; the period is the inverse of frequency. So, if a piano string vibrates at 220 Hz, it means it completes 220 cycles per second. Thus, each cycle takes about \( T = \frac{1}{220} \approx 0.00455 \) seconds.
  • Understanding wave oscillation can help explain sound, light, and even modern technologies like radios.
  • Every wave has a frequency (how many cycles per second) and period (time for one cycle), which are fundamental for grasping wave dynamics.
Piano String Vibration
Piano string vibration is an exemplary case of wave motion and sound production. Each string on the piano is designed to resonate at specific frequencies when struck by a hammer. These vibrations produce sound waves that our ears interpret as musical notes.

Taking the example of the middle A piano note, it resonates at 220 Hz. This means the string vibrates 220 times per second, producing a sound wave with a particular pitch. When you move one octave higher, to an A note sung by a soprano, the frequency doubles to 440 Hz, demonstrating the relationship between frequency and pitch. The vibrations travel as waves through the air, creating what we perceive as sound.
  • This principle of doubling frequency results in what musicians refer to as an octave, which is one of the building blocks of music theory.
  • The physics behind string vibration involves concepts like tension, mass, length, and how they affect the frequency of vibration.

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Most popular questions from this chapter

A small sphere with mass \(m\) is attached to a massless rod of length \(L\) that is pivoted at the top, forming a simple pendulum. The pendulum is pulled to one side so that the rod is at an angle \(\Theta\) from the vertical, and released from rest. (a) In a diagram, show the pendulum just after it is released. Draw vectors representing the forces acting on the small sphere and the acceleration of the sphere. Accuracy counts! At this point, what is the linear acceleration of the sphere? (b) Repeat part (a) for the instant when the pendulum rod is at an angle \(\Theta / 2\) from the vertical. (c) Repeat part (a) for the instant when the pendulum rod is vertical. At this point, what is the linear speed of the sphere?

A \(1.50-\mathrm{kg}\) ball and a \(2.00-\mathrm{kg}\) ball are glued together with the lighter one below the heavier one. The upper ball is attached to a vertical ideal spring of force constant \(165 \mathrm{N} / \mathrm{m},\) and the system is vibrating vertically with amplitude 15.0 \(\mathrm{cm} .\) The gine connecting the balls is old and weak, and it suddenly comes loose when the balls are at the lowest position in their motion. (a) Why is the glue more likely to fail at the lowest point than at any other point in the motion? (b) Find the amplitude and frequency of the vibrations after the lower ball has come loose.

Weighing Astronauts. This procedure has actually been used to "weigh" astronauts in space. A \(42.5-\mathrm{kg}\) chair is attached to a spring and allowed to oscillate. When it is empty, the chair takes 1.30 s to make one complete vibration. But with an astronaut sitting in it, with her feet off the floor, the chair takes 2.54 s for one cycle. What is the mass of the astronaut?

A harmonic oscillator has angular frequency \(\omega\) and amplitude \(A\) . (a) What are the magnitudes of the displacement and velocity when the elastic potential energy is equal to the kinetic energy? (Assume that \(U=0\) at equilibrium, (b) How often does this occur in each cycle? What is the time between occurrences? (c) At an instant when the displacement is equal to \(A / 2,\) what fraction of the total energy of the system is kinetic and what fraction is potential?

If an object on a horizontal, frictionless surface is attached to a spring, displaced, and then released, it will oscillate. If it is displaced 0.120 \(\mathrm{m}\) from its equilibrium position and released with zero initial speed, then after 0.800 \(\mathrm{s}\) its displacement is found to be 0.120 \(\mathrm{m}\) on the opposite side, and it has passed the equilibrium position once during this interval. Find (a) the amplitude; (b) the period; (c) the frequency.

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