A mass is oscillating on the end of a spring. The distance, y, of the mass from its equilibrium point is given by the formula

Question

A mass is oscillating on the end of a spring. The distance, y, of the mass from its equilibrium point is given by the formula $$y=7z\cos(12π\omega t)$$

where y is in centimeters, t is time in seconds, and z and $\omega$ are positive constants. (b) How many oscillations are completed in 1 second?

I don't actually understand this problem. Could someone help me to explain the question and give me some helps about solving the question.

Thanks for the help!!!

Answer 1

One oscillation correspond to the change of $2\pi$ of the argument of the cosine function. So if we start at $t=0$, at time $t=T$, where $T$ is the period of the oscillation, the argument of the cosine changed by $2\pi$. $$12\pi\omega T-12\pi\omega 0=2\pi$$ This yields the period of the oscillation as $$T=\frac{1}{6\omega}$$ Now if you know the period, you can simply calculate how many of those are in one second.

Answer 2

The frequency of oscillation is: $$\nu=\dfrac{\Omega}{2\pi}$$ $$\Omega=12\pi\omega$$ So: $$\nu=6\omega$$ which ise the frequency of oscillation in $Hz$. This means you have $6\omega$ oscillations every second