Question

The equation of motion of a body oscillating on the end of a spring is $frac{d^2x}{dt^2} + 100x = 0$ where $x$ is the displacement in meters of the body from its equilibrium position after time $t$ seconds. Determine $x$ in terms of $t$ given that at time $t = 0$, $x = 2$ and $frac{dx}{dt} = 0.$

          The equation of motion of a body oscillating on the end of a spring is
$frac{d^2x}{dt^2} + 100x = 0$
where $x$ is the displacement in meters of the body from its equilibrium position after time $t$ seconds. Determine $x$ in
terms of $t$ given that at time $t = 0$, $x = 2$ and $frac{dx}{dt} = 0.$

$The equation of motion of a body oscillating on the end of a spring is fracd^2xdt^2 + 100x = 0 where x is the displacement in meters of the body from its equilibrium position after time t seconds. Determine x in terms of t given that at time t = 0, x = 2 and fracdxdt = 0.$

Added by Jose Miguel A.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Scott Stetson

07/04/2024

Step 1

This is a simple harmonic motion equation, where the term $100$ represents the square of the angular frequency $\omega^2$, thus $\omega = \sqrt{100} = 10 \, \text{rad/s}$. Show more…

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The equation of motion of a body oscillating on the end of a spring is: d^2x/dt^2 + 100x = 0 where x is the displacement in meters of the body from its equilibrium position after time t seconds. Determine x in terms of t given that at time t=0, x=2 and dx/dt=0.