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Ch 14: Periodic Motion
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
Chapter 14, Problem 18

A 0.400-kg object undergoing SHM has ax = -1.80 m/s2 when x = 0.300 m. What is the time for one oscillation?

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Start by identifying the formula for acceleration in simple harmonic motion (SHM), which is given by: a = -ω2x, where ω is the angular frequency and x is the displacement.
Substitute the given values into the formula: -1.80 = -ω2×0.300. Solve for ω2.
Calculate ω2 by rearranging the equation: ω2 = 1.800.300.
Find the angular frequency ω by taking the square root of ω2: ω = 1.800.300.
Use the relationship between angular frequency and period: T = ω, where T is the period of one oscillation. Substitute the value of ω to find T.

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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 the restoring force is directly proportional to the displacement and acts in the opposite direction. It is characterized by oscillations around an equilibrium position, such as a mass on a spring or a pendulum.
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Acceleration in SHM

In SHM, acceleration is given by the equation a = -ω²x, where ω is the angular frequency and x is the displacement from the equilibrium position. The negative sign indicates that the acceleration is always directed towards the equilibrium position, opposing the displacement.
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Period of Oscillation

The period of oscillation, T, is the time taken for one complete cycle of motion in SHM. It is related to the angular frequency by the formula T = 2π/ω. Knowing the acceleration and displacement, one can find ω and subsequently calculate the period.
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Related Practice
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 from x = 0.180 m to x = -0.180 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. 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 from x = 0.090 m to x = -0.090 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 0.500kg0.500\(\operatorname{kg}\) 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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