the displacement, d, in millimeters of a tuning fork as a function of time, t, in seconds can be modeled…
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?\n\\(\\frac{1}{3520}\\) hz\n\\(\\frac{1}{1760}\\) hz\n1760 hz\n3520 hz
Answer
Explanation:
Step1: Recall the sine - wave formula
The general form of a sine - wave for simple harmonic motion is $d = A\sin(\omega t)$, where $\omega$ is the angular frequency. In the given equation $d = 0.6\sin(3520\pi t)$, we have $\omega=3520\pi$.
Step2: Use the relationship between angular frequency and frequency
The relationship between angular frequency $\omega$ and frequency $f$ is $\omega = 2\pi f$. We can solve for $f$ by rearranging the formula: $f=\frac{\omega}{2\pi}$. Substitute $\omega = 3520\pi$ into the formula: $f=\frac{3520\pi}{2\pi}$.
Step3: Simplify the expression
Cancel out the $\pi$ terms in the numerator and denominator: $f = 1760$ Hz.
Answer:
1760 Hz