To determine the period of the given function, we need to analyze the characteristics of the sine function within it. The given function is:
[tex]\[ L(t) = \frac{3}{2} \sin \left(\pi t + \frac{\pi}{2}\right) + \frac{3}{2} \][/tex]
Let's recall the general form of a sine function:
[tex]\[ f(t) = A \sin(Bt + C) + D \][/tex]
where:
- [tex]\( A \)[/tex] is the amplitude,
- [tex]\( B \)[/tex] is the coefficient that affects the period,
- [tex]\( C \)[/tex] is the phase shift, and
- [tex]\( D \)[/tex] is the vertical shift.
The period [tex]\( T \)[/tex] of a sine function is determined by the coefficient [tex]\( B \)[/tex] inside the sine function:
[tex]\[ T = \frac{2\pi}{B} \][/tex]
In the given function, we identify that:
[tex]\[ B = \pi \][/tex]
Substituting [tex]\( B = \pi \)[/tex] into the formula for the period, we get:
[tex]\[ T = \frac{2\pi}{\pi} \][/tex]
The [tex]\( \pi \)[/tex] terms cancel out, simplifying to:
[tex]\[ T = 2 \][/tex]
Therefore, the period of the given function [tex]\( L(t) \)[/tex] is [tex]\( 2 \)[/tex] seconds.
Hence, the correct answer is:
2 seconds
