Question

The function$y=A\sin \left(\omega x\right),A>0$has has amplitude 3 and period 2; then$A=\_\_\_\_\_\_\&\omega =\_\_\_\_\_\_.$

Short answer

The function$y=A\sin \left(\omega x\right),A>0$ has amplitude 3 and period 2; then$A=\_3\_\_\&\omega =\_\_\pi \_.$

Step-by-step solution

Step 1. Given Information.

The function$y=A\sin \left(\omega x\right),A>0$has amplitude 3 and period 2; then$A=\_\_\_\_,\omega =\_\_\_\_.$

Step 2. Filling Up.

If $\omega >0$, the amplitude and period of $y=A\sin \left(\omega x\right)$are given by

Amplitude $=\left|A\right|$, Period $T=\frac{2\mathrm{\pi }}{\omega }$.

The amplitude $\left|A\right|=3$.

The Period will be

$$\begin{gathered} 2=\frac{2\mathrm{\pi }}{\omega } \\ \Rightarrow \omega =\pi \end{gathered}$$