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Chapter 11: Problem 23

If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it oscillates. Suppose it is displaced \(0.120 \mathrm{~m}\) from its equilibrium position and released with zero initial speed. 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, and (c) the frequency of the motion.

Short Answer

Expert verified
(a) Amplitude: 0.120 m, (b) Period: 1.600 s, (c) Frequency: 0.625 Hz.

Step by step solution

01

Understand Harmonic Motion

An object attached to a spring and displaced will experience simple harmonic motion. In this situation, the displacement of the object is described by the equation: \[ x(t) = A \cos(\omega t + \phi) \] where \( A \) is the amplitude, \( \omega \) is the angular frequency, \( t \) is time, and \( \phi \) is the phase constant. Since the object is released from rest, \( \phi = 0 \) or \( \phi = \pi \).
02

Determine Amplitude

Amplitude \( A \) is the maximum displacement from the equilibrium position. Given that initially the object is displaced by \( 0.120 \mathrm{~m} \) and then it reaches the same displacement value on the opposite side, the amplitude \( A \) is simply \( 0.120 \mathrm{~m} \).
03

Calculate the Period

The object completes a back and forth motion, crossing the equilibrium point once in the first half of one cycle by reaching the initial amplitude on the opposite side at \( t = 0.800 \) seconds. Thus, \( t = 0.800 \) seconds corresponds to half the period. Therefore, the full period \( T \) is twice this time.\[ T = 2 \times 0.800 \mathrm{~s} = 1.600 \mathrm{~s} \]
04

Find the Frequency

Frequency \( f \) is the reciprocal of the period. Using the period found in the previous step:\[ f = \frac{1}{T} = \frac{1}{1.600 \mathrm{~s}} = 0.625 \mathrm{~Hz} \]

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Key Concepts

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

Amplitude
Amplitude is a key feature of simple harmonic motion. It represents the maximum extent of displacement of an oscillating object from its equilibrium position. In this scenario, the object attached to a spring was initially displaced by **0.120 meters**. This means the amplitude is **0.120 meters**. The amplitude is always positive and directly linked with the energy in the system. More amplitude generally indicates more energy.

When assessing an oscillating system, remember that the oscillation reaches this displacement on both sides around the equilibrium. Since the displacement at both extremes is given and observed, the amplitude is easily determined from such measurements.
Period
The period of a harmonic oscillator is the time taken to complete one full cycle of motion, returning to a particular point in the motion. The motion described occurs over **0.800 seconds** per half of the oscillation, indicating one full oscillation takes **1.600 seconds**.

In broader physics, the period is vital because it remains constant for a given harmonic oscillator system, assuming constant environmental conditions. It gives insights into the timekeeping and synchronization possibilities of oscillatory systems like clocks and metronomes.
Frequency
Frequency refers to how often the oscillation happens in a second. It is the inverse of the period and is measured in **Hertz (Hz)**.In our example, the period is **1.600 seconds**, thus the frequency of oscillation is calculated as follows: \( f = \frac{1}{T} = \frac{1}{1.600 \text{ s}} = 0.625 \text{ Hz} \).

Higher frequency implies more repetitions within a specific time frame. Frequency is essential in understanding not just physical oscillations, but also in fields like electricity, sound waves, and optics.
Harmonic Oscillator
A harmonic oscillator is a system that, once displaced from its equilibrium position, experiences a restoring force proportional to the displacement. This spring-object setup is a classic example. When the spring is stretched or compressed from its original position, it begins to oscillate around the equilibrium point.

Harmonic oscillators can be found in various scientific areas: physics, engineering, and even biology. A simple harmonic oscillator's unique property is that its motion can be described using sinusoidal functions like sine and cosine, which makes mathematical modeling and prediction much simpler.
Angular Frequency
Angular frequency is closely related to both the period and frequency of oscillation.It is represented by \( \omega \) and is defined as the rate of change of the phase of the sinusoidal waveform.

Mathematically, angular frequency is expressed as \( \omega = 2 \pi f \) or \( \omega = \frac{2 \pi}{T} \). It conveys how many radians per second the object oscillates through in its motion.In our scenario, using the frequency calculated, \( \omega = 2 \pi \times 0.625 \text{ Hz} \).

Understanding angular frequency is pivotal, especially in wave mechanics and circular motion, as it allows us to link rotational and linear motion concepts effectively.

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Most popular questions from this chapter

(a) If a vibrating system has total energy \(E_{0},\) what will its total energy be (in terms of \(E_{0}\) ) if you double the amplitude of vibration? (b) If you want to triple the total energy of a vibrating system with amplitude \(A_{0}\), what should its new amplitude be (in terms of \(A_{0}\) )?

A \(0.500 \mathrm{~kg}\) glider on an air track is attached to the end of an ideal spring with force constant \(450 \mathrm{~N} / \mathrm{m} ;\) it undergoes simple harmonic motion with an amplitude of \(0.040 \mathrm{~m}\). Compute (a) the maximum speed of the glider, (b) the speed of the glider when it is at \(x=-0.015 \mathrm{~m},\) (c) the magnitude of the maximum acceleration of the glider, (d) the acceleration of the glider at \(x=-0.015 \mathrm{~m},\) and (e) the total mechanical energy of the glider at any point in its motion.

(a) Music. When a person sings, his or her vocal cords vibrate in a repetitive pattern having the same frequency of the note that is sung. If someone sings the note \(\mathrm{B}\) flat that has a frequency of 466 \(\mathrm{Hz}\), 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 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?

A simple pendulum in a science museum entry hall is \(3.50 \mathrm{~m}\) long, has a \(1.25 \mathrm{~kg}\) bob at its lower end, and swings with an amplitude of \(11.0^{\circ} .\) How much time does the pendulum take to swing from its extreme right side to its extreme left side?

A steel wire has the following properties: $$ \begin{array}{l} \text { Length }=5.00 \mathrm{~m} \\ \text { Cross-sectional area }=0.040 \mathrm{~cm}^{2} \end{array} $$ Young's modulus \(=2.0 \times 10^{11} \mathrm{~Pa}\) Shear modulus \(=0.84 \times 10^{11} \mathrm{~Pa}\) Proportional limit \(=3.60 \times 10^{8} \mathrm{~Pa}\) Breaking stress \(=11.0 \times 10^{8} \mathrm{~Pa}\) The proportional limit is the maximum stress for which the wire still obeys Hooke's law (see point \(\mathrm{B}\) in Figure 11.12 ). The wire is fastened at its upper end and hangs vertically. (a) How great a weight can be hung from the wire without exceeding the proportional limit? (b) How much does the wire stretch under this load? (c) What is the maximum weight that can be supported?
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