Question

problem 2:
a mass attached to a spring oscillates back and forth in the x - direction,
given by ( x=(6 mathrm{~cm}) cos (5.84 t) ), where ( t ) is in seconds. give the amplitude
and frequency of the oscillations. plot the position as a function of time.

Explanation:

Step1: Find the amplitude

The general form of the cosine function for simple harmonic motion is \(x = A\cos(\omega t)\), where \(A\) is the amplitude.
Comparing \(x=(6\ cm)\cos(5.84t)\) with \(x = A\cos(\omega t)\), we get \(A = 6\ cm\).

Step2: Find the frequency

We know that \(\omega=2\pi f\), where \(\omega\) is the angular frequency and \(f\) is the frequency.
Given \(\omega = 5.84\ rad/s\), then \(f=\frac{\omega}{2\pi}\).
Substitute \(\omega = 5.84\) into the formula: \(f=\frac{5.84}{2\pi}\ s^{-1}\approx0.93\ Hz\).

Answer:

Amplitude \(A = 6\ cm\), Frequency \(f\approx0.93\ Hz\)