Answer:
a) [tex]f = 1.642\,hz[/tex], b) [tex]\bar x = 0.095\,m[/tex], c) [tex]A = 0.374\,m[/tex], d) [tex]v_{max} = 3.859\,\frac{m}{s}[/tex], e) [tex]a_{max} = 39.809\,\frac{m}{s^{2}}[/tex], f) [tex]k = 24.162\,\frac{N}{m}[/tex], g) [tex]E = 1.690\,J[/tex]
Explanation:
a) The frequency is the reciprocal of the period of oscillation:
[tex]f = \frac{1}{T}[/tex]
[tex]f = \frac{1}{0.609\,s}[/tex]
[tex]f = 1.642\,hz[/tex]
b) The equilibrium position is the average of the extreme points:
[tex]\bar x = \frac{-0.279\,m +0.469\,m}{2}[/tex]
[tex]\bar x = 0.095\,m[/tex]
c) The amplitude is the absolute of the substraction of the equilibrium position from any of the extreme points:
[tex]A = |0.469\,m - 0.095\,m|[/tex]
[tex]A = 0.374\,m[/tex]
d) The angular frequency is:
[tex]\omega = 2\pi \cdot f[/tex]
[tex]\omega = 2\pi \cdot (1.642\,hz)[/tex]
[tex]\omega \approx 10.317\,\frac{rad}{s}[/tex]
The maximum speed is:
[tex]v_{max} = \omega \cdot A[/tex]
[tex]v_{max} = (10.317\,\frac{rad}{s} )\cdot (0.374\,m)[/tex]
[tex]v_{max} = 3.859\,\frac{m}{s}[/tex]
e) The maximum acceleration is:
[tex]a_{max} = \omega^{2}\cdot A[/tex]
[tex]a_{max} = (10.317\,\frac{rad}{s} )^{2}\cdot (0.374\,m)[/tex]
[tex]a_{max} = 39.809\,\frac{m}{s^{2}}[/tex]
f) The force constant is:
[tex]k = \omega^{2}\cdot m[/tex]
[tex]k = (10.317\,\frac{rad}{s} )^{2}\cdot (0.227\,kg)[/tex]
[tex]k = 24.162\,\frac{N}{m}[/tex]
g) The total mechanical energy is:
[tex]E = \frac{1}{2}\cdot k \cdot A^{2}[/tex]
[tex]E = \frac{1}{2}\cdot (24.162\,\frac{N}{m} )\cdot (0.374\,m)^{2}[/tex]
[tex]E = 1.690\,J[/tex]
