Question

When represented on an oscilloscope, the note A above middle C has

period of $\frac{1}{440}$. Which of the following can be an equation for an oscilloscope graph of this note? The amplitude of the graph is $K$.

a. $y=K\sin 220\pi t$

b. $y=K\sin 440\pi t$

c. $y=K\sin 880\pi t$

Short answer

The answer is c, $y=K\sin 880\pi t$

Step-by-step solution

Step 1. Given information

Three options are given.

Step 2. Concept used

The only difference in the options is angular frequency. So, find the angular frequency of the period.

Step 3. Calculation

The angular frequency is-

$\begin{array}{l}\omega =\frac{2\pi }{t}\\ \omega =\frac{2\pi }{\frac{1}{440}}\\ \omega =880\pi \end{array}$

So the equation is,

$y=K\sin 880\pi t$