Question

the displacement, d, in millimeters of a tuning fork as a function of time, t, in seconds can be modeled with the equation d = 0.6sin(3520πt). what is the frequency of the tuning fork?
1/3520 hz
1/1760 hz
1760 hz
3520 hz

Explanation:

Step1: Recall the general form of a sinusoidal - wave equation

The general form of a sinusoidal - wave equation for displacement is $d = A\sin(\omega t)$, where $\omega$ is the angular frequency. In the given equation $d = 0.6\sin(3520\pi t)$, the angular frequency $\omega=3520\pi$.

Step2: Use the relationship between angular frequency and frequency

The relationship between angular frequency $\omega$ and frequency $f$ is $\omega = 2\pi f$. We need to solve for $f$. Rearranging the formula $f=\frac{\omega}{2\pi}$.

Step3: Substitute the value of $\omega$ into the formula for $f$

Substitute $\omega = 3520\pi$ into $f=\frac{\omega}{2\pi}$. Then $f=\frac{3520\pi}{2\pi}$.

Step4: Simplify the expression

Canceling out the $\pi$ terms, we get $f = 1760$ Hz.

Answer:

C. 1760 Hz