Problem
Submit quiz function over two periods. \[ I(t)=120 \sin \left(20 \pi t-\frac{\pi}{4}\right), t \geq 0 \] The period is $\square$. (Simplify your answer. Type an exact answer, using $\pi$ as needed. Use integers or fractions for any numbers in the expression.) The amplitude is $\square$. (Simplify your answer. Type an exact answer, using $\pi$ as needed. Use integers or fractions for any numbers in the expression.) The phase shift is $\square$. (Simplify your answer. Type an exact answer, using $\mathrm{z}$ as needed. Use integers or fractions for any numbers in the expression) Choose the correct graph below. A. B. c. D. Next
Solution
Step 1 ：The function is of the form \(I(t) = A \sin(Bt - C)\), where \(A\) is the amplitude, \(B\) determines the period, and \(C\) is the phase shift.
Step 2 ：To find the period, we use the formula \(2\pi / |B|\). In this case, \(B = 20\pi\), so the period is \(2\pi / |20\pi| = \frac{1}{10}\).
Step 3 ：The amplitude is the absolute value of \(A\), which in this case is \(120\).
Step 4 ：The phase shift is given by \(C / B\). In this case, \(C = -\pi/4\) and \(B = 20\pi\), so the phase shift is \(-\pi/4 / 20\pi = -\frac{1}{80}\).
Step 5 ：Final Answer: The period is \(\boxed{\frac{1}{10}}\), the amplitude is \(\boxed{120}\), and the phase shift is \(\boxed{-\frac{1}{80}}\).
