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Ch 14: Periodic Motion
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 14th EditionCh 14: Periodic MotionProblem 24b
Chapter 14, Problem 24b

For the oscillating object in Fig. E14.4, what is its maximum acceleration?
Graph depicting position over time for a simple harmonic oscillator, labeled with time intervals.

Verified step by step guidance
1
Identify the amplitude of the oscillation from the graph. The amplitude is the maximum displacement from the equilibrium position, which is 10 cm in this case.
Determine the angular frequency \( \omega \) of the oscillation. The period \( T \) can be found from the graph as the time it takes to complete one full cycle. From the graph, the period \( T \) is 10 seconds. Use the formula \( \omega = \frac{2\pi}{T} \) to calculate the angular frequency.
Recall the formula for maximum acceleration in simple harmonic motion: \( a_{max} = \omega^2 A \), where \( A \) is the amplitude and \( \omega \) is the angular frequency.
Substitute the values of \( \omega \) and \( A \) into the formula for maximum acceleration to find \( a_{max} \).
Ensure the units are consistent. Since the amplitude is given in cm, convert it to meters by dividing by 100 if necessary, to ensure the acceleration is in \( \text{m/s}^2 \).

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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 position, velocity, and acceleration graphs. In SHM, the maximum displacement from the equilibrium is known as the amplitude.
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Acceleration in SHM

In Simple Harmonic Motion, the acceleration of the oscillating object is directly related to its displacement from the equilibrium position. The maximum acceleration occurs at the maximum displacement (amplitude) and is given by the formula a_max = ω²A, where ω is the angular frequency and A is the amplitude. This relationship highlights how acceleration varies throughout the oscillation.
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Graph Interpretation

The graph provided depicts the position of the oscillating object over time, illustrating its periodic nature. By analyzing the graph, one can determine key parameters such as amplitude, period, and phase. The maximum displacement from the horizontal axis indicates the amplitude, while the steepest slopes correspond to maximum velocity, and the points of zero slope indicate maximum acceleration.
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Related Practice
Textbook Question

A 0.500-kg glider, attached to the end of an ideal spring with force constant k = 450 N/m, undergoes SHM with an amplitude of 0.040 m. Compute the speed of the glider when it is at x = -0.015 m.

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Textbook Question

For the oscillating object in Fig. E14.4, what is its maximum speed?

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Textbook Question

A 0.500kg0.500\(\operatorname{kg}\) mass on a spring has velocity as a function of time given by vx(t)=−(3.60cm/s)sin[(4.7 rad/s)t−(π/2)]v_{x}(t)=-(3.60\(\operatorname{cm}\)/s)\(\sin\)[(4.7\(\text{ }\)rad/s)t-(\(\pi\)/2)]. What are the period and the force constant of the spring?

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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. The amplitude of the motion is 0.165 m. The maximum speed of the block is 3.90 m/s. What is the maximum magnitude of the acceleration of the block?

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Textbook Question

A 0.500kg0.500\(\operatorname{kg}\) mass on a spring has velocity as a function of time given by vx(t)=−(3.60cm/s)sin[(4.7 rad/s)t−(π/2)]v_{x}(t)=-(3.60\(\operatorname{cm}\)/s)\(\sin\)[(4.7\(\text{ }\)rad/s)t-(\(\pi\)/2)]. What are the amplitude and the maximum acceleration of the mass?

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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. The amplitude of the motion is 0.250 m and the period is 3.20 s. What are the speed and acceleration of the block when x = 0.160 m?

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