The displacement of a mass from its equilibrium position as it oscillates on a horizontal spring satisfies the equation \(x(t)= Acos(\omega t + \phi_0)\).

At \(t = 0 s\), if \( x = - \frac{A}{2} \), what are the two possible values of \(\phi_0\) between 0 and 2\(\pi\) radians?

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The displacement of a mass undergoing simple harmonic motion from its equilibrium position can be expressed as $x(t) = A \cos(\omega t + \phi_0)$ where $A$ is the amplitude of the oscillation (the magnitude of the largest displacement from equilibrium); and $\phi(t) = \omega t + \phi_0$ is the phase, expressed in (dimensionless) radians. If one thinks of the displacement from equilibrium as the projection of uniform circular motion on the $x$-axis (or $x$-component), $\phi(t)$ represents the angle measured counterclockwise from the $x$-axis as seen in the Figure. $\omega = rac{2\pi}{T}$ is the angular frequency (in rad/s), $T$ is the period (in seconds), and $\phi_0$ is the value of the phase at $t = 0$ s, also known as the phase constant.

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Attribution

Problem is from the OpenStax University Physics Volume 2 textbook, licensed under the CC-BY 4.0 license.