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Ch 14: Periodic Motion
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Non è quello che usi tu?Cambia libro di testo
Capitolo 14, Problema 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 = 2πω, where T is the period of one oscillation. Substitute the value of ω to find T.

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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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