To find the period of the simple harmonic motion described by the equation [tex]\( d = 2 \sin \left(\frac{\pi}{3} t\right) \)[/tex], we need to identify the angular frequency [tex]\( \omega \)[/tex] and use it to determine the period [tex]\( T \)[/tex].
1. Identify the form of the equation:
The given equation is [tex]\( d = 2 \sin \left(\frac{\pi}{3} t\right) \)[/tex], which matches the general form for simple harmonic motion:
[tex]\[ d = A \sin(\omega t) \][/tex]
where [tex]\( A \)[/tex] is the amplitude, and [tex]\( \omega \)[/tex] is the angular frequency.
2. Determine the angular frequency [tex]\( \omega \)[/tex]:
From the equation [tex]\( d = 2 \sin \left(\frac{\pi}{3} t\right) \)[/tex], we can see that the angular frequency [tex]\( \omega \)[/tex] is:
[tex]\[ \omega = \frac{\pi}{3} \][/tex]
3. Use the relationship between the angular frequency [tex]\( \omega \)[/tex] and the period [tex]\( T \)[/tex]:
The period [tex]\( T \)[/tex] of simple harmonic motion is given by the formula:
[tex]\[ T = \frac{2\pi}{\omega} \][/tex]
4. Substitute [tex]\( \omega = \frac{\pi}{3} \)[/tex] into the formula for the period:
[tex]\[ T = \frac{2\pi}{\frac{\pi}{3}} \][/tex]
5. Simplify the expression:
[tex]\[ T = \frac{2\pi}{\frac{\pi}{3}} = 2\pi \cdot \frac{3}{\pi} = 6 \][/tex]
Thus, the period [tex]\( T \)[/tex] of the simple harmonic motion described by the equation [tex]\( d = 2 \sin \left(\frac{\pi}{3} t\right) \)[/tex] is 6.
So, the period [tex]\( T \)[/tex] is [tex]\( \boxed{6} \)[/tex].
