To find the period of the simple harmonic motion described by the equation [tex]\( d = 5 \sin (2 \pi t) \)[/tex], we need to identify the coefficient of [tex]\( t \)[/tex] inside the sine function and use the relationship between this coefficient and the period of the function.
The general form for simple harmonic motion is [tex]\( d = A \sin (Bt) \)[/tex], where:
- [tex]\( A \)[/tex] is the amplitude,
- [tex]\( B \)[/tex] is the angular frequency.
The period [tex]\( T \)[/tex] of the function [tex]\( \sin (Bt) \)[/tex] is given by the formula:
[tex]\[ T = \frac{2\pi}{B} \][/tex]
For the given equation [tex]\( d = 5 \sin (2 \pi t) \)[/tex]:
- The amplitude [tex]\( A \)[/tex] is 5 (though this does not affect the period).
- The angular frequency [tex]\( B \)[/tex] is [tex]\( 2\pi \)[/tex].
Substitute [tex]\( B = 2\pi \)[/tex] into the period formula:
[tex]\[ T = \frac{2\pi}{2\pi} \][/tex]
Simplify the fraction:
[tex]\[ T = 1 \][/tex]
So, the period [tex]\( T \)[/tex] of the simple harmonic motion described by the equation [tex]\( d = 5 \sin (2 \pi t) \)[/tex] is 1.
