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Ch 14: Periodic Motion
All textbooksYoung & Freedman Calc 14th EditionCh 14: Periodic MotionProblem 14
Chapter 14, Problem 14

A 0.400-kg object undergoing SHM has ax = -1.80 m/s^2 when x = 0.300 m. What is the time for one oscillation?

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Identify the given values: mass (m) = 0.400 kg, acceleration (a_x) = -1.80 m/s^2, and displacement (x) = 0.300 m.
Use the formula for acceleration in simple harmonic motion (SHM), which is a_x = -\omega^2 x, where \omega is the angular frequency. Rearrange the formula to solve for \omega: \omega = \sqrt{-a_x / x}.
Calculate the angular frequency (\omega) using the values of a_x and x.
Determine the period of oscillation (T), which is the time for one complete oscillation. The period T is related to angular frequency by the formula T = 2\pi / \omega.
Calculate the period (T) using the angular frequency (\omega) obtained in the previous step.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Simple Harmonic Motion (SHM)

Simple Harmonic Motion is a type of periodic motion where an object oscillates around an equilibrium position. The motion is characterized by a restoring force proportional to the displacement from the equilibrium, leading to sinusoidal motion. In SHM, the acceleration is always directed towards the equilibrium position and is given by the formula a = -ω²x, where ω is the angular frequency.
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Acceleration in SHM

In Simple Harmonic Motion, the acceleration of the object is directly related to its displacement from the equilibrium position. The formula a = -ω²x indicates that the acceleration is negative when the object is displaced positively, meaning it acts to restore the object back to equilibrium. This relationship is crucial for determining the properties of the oscillation, including the angular frequency.
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Period of Oscillation

The period of oscillation in SHM is the time taken for one complete cycle of motion. It is denoted by T and is inversely related to the frequency of the motion. The period can be calculated using the formula T = 2π/ω, where ω is the angular frequency. Understanding the period is essential for analyzing the time characteristics of oscillatory systems.
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Related Practice
Textbook Question
The point of the needle of a sewing machine moves in SHM along the x-axis with a frequency of 2.5 Hz. At t = 0 its position and velocity components are +1.1 cm and -15 cm/s, respectively. (a) Find the acceleration component of the needle at t = 0.
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Textbook Question
A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the amplitude of the motion is 0.090 m, it takes the block 2.70 s to travel from x = 0.090 m to x = -0.090 m. If the amplitude is doubled, to 0.180 m, how long does it take the block to travel (a) from x = 0.180 m to x = -0.180 m?
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Textbook Question
Weighing Astronauts. This procedure has been used to 'weigh' astronauts in space: A 42.5-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?
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Textbook Question
A 0.500-kg mass on a spring has velocity as a function of time given by vx(t) = -(3.60 cm/s) sin[(4.71 rad/s)t - (pi/2)]. What are (a) the period; (b) the amplitude; (c) the maximum acceleration of the mass; (d) the force constant of the spring?
1574
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Textbook Question
A 0.500-kg mass on a spring has velocity as a function of time given by vx(t) = -(3.60 cm/s) sin[(4.71 rad/s)t - (pi/2)]. What are (a) the period; (b) the amplitude; (c) the maximum acceleration of the mass; (d) the force constant of the spring?
807
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Textbook Question
You pull a simple pendulum 0.240 m long to the side through an angle of 3.50° and release it. (a) How much time does it take the pendulum bob to reach its highest speed?
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