What Are the Characteristics of Oscillations for a Box on a Spring?

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The discussion focuses on the oscillatory motion of a box on a spring, described by the equation x = 5.3 m * cos(20/sec * t). Key calculations include determining the position of the box after 2 seconds, which is approximately 5.28 m, the amplitude of oscillation at 5.3 m, and the angular frequency (ω) of 20 s-1. The period of oscillation can be calculated as T = 2π/ω, resulting in a period of approximately 0.314 seconds. Maximum velocity and acceleration can also be derived from the amplitude and angular frequency.

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  • Calculate the period of oscillation using the formula T = 2π/ω
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the position of the center of the box shown is given by the equation

x = 5.3 m * cos(20/sec * t)
(a) What is the position of the box 2 seconds after the oscillations have started?

(b) What is the amplitude of the box's oscillations?

(c) What is the period of the box's oscillations?

(d) What is the box's maximum velocity?

(e) What is the box's maximum acceleration?

(f) How long does it take the box to move from -2.65 m to +2.65 m?

for a, i just plugged in 2 for t and got 4.98...but that's wrong not sure where to go from there
Calc:
5.3*cos(20/(1/cos(2)))...sec= 1/cos
 
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Start with x = A cos (ωt). What is A? What is ω?

A = 5.3 m, and ω = 20 s-1
 


so x=5.3*cos(20E-1*2)...5.28?
 

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