Answer:
The period of this pendulum is [tex]T=1.55\: s[/tex]
Explanation:
The equation of motion of a pendulum is given by:
[tex]\frac{d\theta^{2}}{dt^{2}}+\frac{g}{L}sin(\theta)=0[/tex] (1)
Where:
θ is the angle of motion
g is the gravity at the earth surface (9.81 m/s²)
L is the length of the pendulum (0.6 m)
Now, using equation (1) we can find the square angular frequency (ω), it will be:
[tex]\omega^{2}=\frac{g}{L}[/tex]
[tex]\omega=\sqrt{\frac{g}{L}}[/tex]
Let's recall that the angular frequency is [tex]\omega=\frac{2\pi}{T}[/tex], then the period will be:
[tex]T=\frac{2\pi}{\omega}[/tex]
[tex]T={2\pi}\sqrt{\frac{L}{g}[/tex]
[tex]T=2\pi}\sqrt{\frac{0.6}{9.81}[/tex]
Therefore, the period of this pendulum is [tex]T=1.55\: s[/tex]
I hope it helps you!
