Answer:
a) 2.67 m/s
b) 0.82 J
Explanation:
Amplitude A = 25 cm = 0.25 m
The period of the motion is the inverse of the frequency
[tex]T = \frac{1}{f} = \frac{1}{1.7} = 0.588 s[/tex]
So the angular frequency
[tex]\omega = \frac{2\pi}{T} = \frac{2\pi}{0.588} = 10.68 rad/s[/tex]
The speed at the equilibrium point is the maximum speed, at
[tex]v = \omega A = 10.68 * 0.25 = 2.67 m/s[/tex]
The spring constant can be calculated using the following
[tex]\omega^2 = \frac{k}{m} = \frac{k}{0.23}[/tex]
[tex]k = 0.23\omega^2 = 0.23*10.68^2 = 26.24 N/m[/tex]
The total energy of the oscillation is
[tex]E = kA^2 / 2 = 26.24*0.25^2 / 2 = 0.82 J[/tex]
