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A \(0.400-\mathrm{kg}\) object undergoing \(\mathrm{SHM}\) has \(a_{x}=-2.70 \mathrm{m} / \mathrm{s}^{2}\) when \(x=0.300 \mathrm{m}\) . What is the time for one oscillation?

Short Answer

Expert verified
The time for one oscillation is approximately 2.094 seconds.

Step by step solution

01

Understand the relationship in SHM

In Simple Harmonic Motion (SHM), the acceleration is related to the displacement by the formula \( a = -\omega^2 x \), where \( \omega \) is the angular frequency and \( x \) is the displacement.
02

Solve for angular frequency

We have the acceleration \( a = -2.70 \, \text{m/s}^2 \) and the displacement \( x = 0.300 \, \text{m} \). Plug these values into the formula: \( -2.70 = -\omega^2 \times 0.300 \). Solving for \( \omega^2 \) gives \( \omega^2 = \frac{2.70}{0.300} = 9 \). Therefore, \( \omega = 3 \, \text{rad/s} \).
03

Calculate the period

The period \( T \) of SHM is given by \( T = \frac{2\pi}{\omega} \). Substituting \( \omega = 3 \, \text{rad/s} \), we get \( T = \frac{2\pi}{3} \approx 2.094 \, \text{s} \).

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Key Concepts

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

Angular Frequency
In Simple Harmonic Motion (SHM), angular frequency is a critical parameter. It tells us how fast an object oscillates, moving back and forth over time. Angular frequency, denoted by \( \omega \), is linked to the physical properties of the system, such as the mass and the stiffness of the "spring" that is causing the oscillation.
The formula to determine angular frequency is \( \omega = \sqrt{\frac{k}{m}} \), where \( k \) is the spring constant and \( m \) is the mass of the object.
  • In the given exercise, we calculated \( \omega \) using the specific acceleration and displacement, rearranging the equation \( a = -\omega^2 \cdot x \).
  • By solving \( -2.70 = -\omega^2 \times 0.300 \), we derived \( \omega = 3 \, \text{rad/s} \).
This measurement is in radians per second (rad/s), which indicates the speed of oscillation in a circular manner even if the path is linear. Understanding \( \omega \) helps predict how the system behaves over time.
Period of Oscillation
The period of oscillation, denoted by \( T \), is another key aspect of SHM. It speaks to the time it takes for the object to complete one full cycle of motion. This concept is crucial for understanding how long an oscillatory motion takes from start to finish.
The relationship between angular frequency and the period is given by the formula \( T = \frac{2\pi}{\omega} \). This equation highlights how the period becomes shorter as angular frequency increases.
  • In the exercise, with an angular frequency of \( 3 \, \text{rad/s} \), we computed \( T = \frac{2\pi}{3} \).
  • The result is approximately \( 2.094 \, \text{s} \), meaning it takes just over 2 seconds for the object to return to its initial position.
Knowing the period provides insights into the dynamics of the oscillating system, especially important in engineering and physics for designing systems that involve periodic motion.
Acceleration in SHM
Acceleration in Simple Harmonic Motion (SHM) is fascinating as it varies with the position of the oscillating object. It is directly proportional to displacement, meaning that as the object moves further from the central point, the acceleration increases but in the opposite direction.
The equation \( a = -\omega^2 x \) expresses this behavior, where \( a \) stands for acceleration, \( \omega \) is angular frequency, and \( x \) is displacement.
  • In this exercise, the acceleration was given as \( -2.70 \, \text{m/s}^2 \) when \( x = 0.300 \, \text{m} \).
  • This negative sign indicates that the acceleration acts in the opposite direction to displacement, which is characteristic of SHM.
Acceleration in SHM ensures the system remains in stable oscillation, causing the object to decelerate as it moves away and accelerate as it returns. This continuous interplay leads to a smooth periodic motion, essential for systems ranging from clocks to seismic sensors.

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Most popular questions from this chapter

A building in San Francisco has light fixtures consisting of small \(2.35-\mathrm{kg}\) bulbs with shades hanging from the ceiling at the end of light thin cords \(1.50 \mathrm{~m}\) long. If a minor earthquake occurs, how many swings per second will these fixtures make?

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 sinusoidally varying driving force is applied to a damped harmonic oscillator of force constant \(k\) and mass \(m\) . If the damping constant has a value \(b_{1}\) , the amplitude is \(A_{1}\) when the driving angular frequency equals \(\sqrt{k / m} .\) In terms of \(A_{1},\) what is the amplitude for the same driving frequency and the same driving force amplitude \(F_{\text { max }}\) if the damping constant is \((a) 3 b_{1}\) and \((b) b_{1} / 2 ?\)

A \(1.50-\mathrm{kg}\) mass on a spring has displacement as a function of time given by the equation $$x(t)=(7.40 \mathrm{cm}) \cos \left[\left(4.16 \mathrm{s}^{-1}\right) t-2.42\right]$$ Find (a) the time for one complete vibration; (b) the force constant of the spring; (c) the maximum speed of the mass; (d) the maximum force on the mass; (e) the position, speed, and acceleration of the mass at \(t=1.00 \mathrm{s} ;(\mathrm{f})\) the force on the mass at that time.

A mass \(m\) is attached to one end of a massess spring with a force constant \(k\) and an unstretched length \(l_{0}\) . The other end of the spring is free to turn about a nail driven into a frictionless, horizontal surface (Fig. 13.44\() .\) The mass is made to revolve in a circle with an angular frequency of revolution \(\omega^{\prime}\) , (a) Calculate the length \(l\) of the spring as a function of \(\omega^{\prime} .\) (b) What happens to the result in part (a) when \(\omega^{\prime}\) approaches the natural frequency \(\omega=\sqrt{k} / m\) of the mass-spring system? (ff your result bothers you, remember that massless springs and frictionless surfaces don't exist as such, but are only approximate descriptions of real springs and surfaces. Also, Hooke's law is only an approximation of the way real springs behave; the greater the elongation of the spring, the greater the deviation from Hooke's law.)

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