To determine the period of the simple harmonic motion described by the equation [tex]\( d = 5 \sin(2t) \)[/tex], we need to understand how the period is related to the equation of motion.
The general form of a simple harmonic motion equation is:
[tex]\[ d = A \sin(Bt + C) + D \][/tex]
Where:
- [tex]\( A \)[/tex] is the amplitude of the motion.
- [tex]\( B \)[/tex] is the angular frequency (in radians per unit time).
- [tex]\( t \)[/tex] is the time variable.
- [tex]\( C \)[/tex] is the phase shift.
- [tex]\( D \)[/tex] is the vertical shift.
For our specific equation [tex]\( d = 5 \sin(2t) \)[/tex], we can identify the following parameters:
- The coefficient of the sine function inside the trigonometric function is 2. This coefficient represents the angular frequency [tex]\( B \)[/tex].
The period [tex]\( T \)[/tex] of a simple harmonic motion is determined by the angular frequency [tex]\( B \)[/tex], and it can be found using the formula:
[tex]\[ T = \frac{2\pi}{B} \][/tex]
In our case, [tex]\( B = 2 \)[/tex]. Substituting this value into the period formula, we get:
[tex]\[ T = \frac{2\pi}{2} \][/tex]
Simplifying this expression:
[tex]\[ T = \pi \][/tex]
Therefore, the period of the simple harmonic motion described by the equation [tex]\( d = 5 \sin(2t) \)[/tex] is [tex]\( \pi \)[/tex] units of time.
