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A harmonic oscillator consists of a \(0.500-\mathrm{kg}\) mass attached to an ideal spring with force constant 140 \(\mathrm{N} / \mathrm{m}\) . Find (a) the period; (b) the frequency; (c) the angular frequency of the oscillations.

Short Answer

Expert verified
Period: 0.375 s; Frequency: 2.67 Hz; Angular frequency: 16.79 rad/s.

Step by step solution

01

Identifying Key Information

The mass of the oscillator is given as \( m = 0.500 \, \text{kg} \) and the force constant of the spring is \( k = 140 \, \text{N/m} \). We need to use these to find the period, frequency, and angular frequency.
02

Calculate the Period (T)

The period \( T \) of a harmonic oscillator can be found using the formula \[ T = 2\pi \sqrt{\frac{m}{k}} \]. Substituting \( m = 0.500 \) kg and \( k = 140 \) N/m, we get: \[ T = 2\pi \sqrt{\frac{0.500}{140}} \approx 2\pi \sqrt{0.00357} \approx 2\pi (0.0597) \approx 0.375 \, \text{seconds} \].
03

Calculate the Frequency (f)

The frequency \( f \) is the reciprocal of the period, given by \( f = \frac{1}{T} \). Using \( T = 0.375 \) seconds from Step 2, we find: \[ f \approx \frac{1}{0.375} \approx 2.67 \, \text{Hz} \].
04

Calculate the Angular Frequency (ω)

The angular frequency \( \omega \) can be calculated using the formula \( \omega = 2\pi f \). Substituting \( f = 2.67 \, \text{Hz} \), we get: \[ \omega = 2\pi \times 2.67 \approx 16.79 \, \text{rad/s} \].

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Key Concepts

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

Period of Oscillation
In a harmonic oscillator, the period of oscillation is the time it takes to complete one full cycle of motion. For a mass-spring system, the period is determined by the properties of the mass and the spring constant. The formula to calculate the period \( T \) is given by:
  • \[ T = 2\pi \sqrt{\frac{m}{k}} \]
Here, \( m \) is the mass attached to the spring, and \( k \) is the spring constant, measuring how stiff the spring is.
Substituting known values, the formula helps us understand how the system’s mass and the spring's force constant affects the oscillation time. A larger mass or a less stiff spring increases the period, meaning the system takes longer to complete an oscillation.
Frequency Calculation
The frequency of a harmonic oscillator tells us how many complete oscillations occur in a unit time, typically a second. Once you know the period \( T \), finding the frequency is straightforward.
The relationship between frequency \( f \) and period is:
  • \[ f = \frac{1}{T} \]
This inverse relationship means a shorter period leads to a higher frequency, implying the system oscillates more times per second. In our example, using the calculated period, the frequency was found to be approximately \( 2.67 \text{ Hz} \), meaning the mass completes about 2.67 cycles every second. This insight is critical in applications needing precise timing and control of oscillations.
Angular Frequency
Angular frequency gives us a different perspective on the oscillation's speed. It describes how fast the oscillator moves through each angle in its cycle, rather than counting full cycles per second.
For a harmonic oscillator, angular frequency \( \omega \) is related to the regular frequency by the formula:
  • \[ \omega = 2\pi f \]
Substituting our frequency value, we arrive at an angular frequency of around \( 16.79 \text{ rad/s} \).
This measure is particularly useful in physics and engineering, as it directly ties into equations of motion and energy calculations, using radians as a natural measure of rotational cycle.
Spring Constant
The spring constant \( k \), or force constant, is a fundamental property of the spring in the harmonic oscillator. It quantifies the spring's stiffness or resistance to deformation. The larger the spring constant, the stiffer the spring.
The spring constant impacts the period of oscillation significantly. It enters the period equation \( T = 2\pi \sqrt{\frac{m}{k}} \) under the square root, illustrating how an increase in stiffness (a larger \( k \)) will decrease the period. Thus, stiff springs lead to faster oscillations and a shorter time for each cycle. Understanding this relationship helps design systems for specific oscillation characteristics by choosing an appropriate spring constant.
Mass-Spring System
A mass-spring system is a classic physical representation of a harmonic oscillator. It consists of a mass attached to a spring that can stretch and compress.
Key factors affecting the oscillation include:
  • Mass \( m \): Heavier masses increase the period and lower the frequency, slowing down oscillations.
  • Spring constant \( k \): A stiffer spring (higher \( k \)) reduces the period and increases the frequency.
This system is governed by Hooke's Law, where the spring force is proportional to the displacement, making it a linear oscillator with predictable motion.
Analyzing such systems is essential in engineering and physics as they model real-world structures like bridges, vehicle suspensions, and various electronic components.

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Most popular questions from this chapter

A rocket is accelerating upward at 4.00 \(\mathrm{m} / \mathrm{s}^{2}\) from the launchpad on the earth. Inside a small \(1.50-\mathrm{kg}\) ball hangs from the ceiling by a light \(1.10-\mathrm{m}\) wire. If the ball is displaced \(8.50^{\circ}\) from the vertical and released, find the amplitude and period of the resulting swings of this pendulum.

When displaced from equilibrium, the two liydrogen atoms in an \(\mathrm{H}_{2}\) molecule are acted on by a restoring force \(F_{x}=-k x\) with \(k=580 \mathrm{N} / \mathrm{m}\) . Calculate the oscillation frequency of the \(\mathrm{H}_{2}\) molecule. (Hint: The mass of a hydrogen atom is 1.008 atomic mass units, or 1 u; see Appendix E. As in Example 13.7 (Section \(13.4 ),\) use \(m / 2\) instead of \(m\) in the expression for \(f\) )

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A 5.00 kg partridge is suspended from a pear tree by an ideal spring of negligible mass. When the partridge is pulled down 0.100 \(\mathrm{m}\) below its equilibrium position and released, it vibrates with a period of 4.20 \(\mathrm{s}\) . (a) What is its speed as it passes through the equilibrium position? (b) What is its acceleration when it is 0.050 \(\mathrm{m}\) above the equilibrium position? (c) When it is moving upward, how much time is required for it to move from a point 0.050 \(\mathrm{m}\) below its equilibrium position to a point 0.050 \(\mathrm{m}\) above it? (d) The motion of the partridge is stopped, and then it is removed from the spring. How much does the spring shorten?

A cheerleader waves her pom-pom in SHM with an amplitude of 18.0 \(\mathrm{cm}\) and a frequency of 0.850 \(\mathrm{Hz}\) . Find (a) the maximum magnitude of the acceleration and of the velocity; \((6)\) the acceleration and speed when the poin-poin's coordinate is \(x=+9.0 \mathrm{cm}\); (c) the time required to move from the equilibrium position directly to a point 12.0 \(\mathrm{cm}\) away. (d) Which of the quantities asproach used parts \((\mathrm{a}),(\mathrm{b}),\) and \((\mathrm{c})\) can be found using the energy approach used in Section \(13.3,\) and which cannot? Explain.

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