Answer:
a) [tex]T = 3.097\,s[/tex], b) [tex]v_{max} = 3.449\times 10^{3}\,\frac{m}{s}[/tex], c) [tex]E = 5.477\times 10^{-7}\,J[/tex], d) [tex]F = 6.443\times 10^{-4}\,N[/tex], e) [tex]F = 3.222\times 10^{-4}\,N[/tex]
Explanation:
a) The angular frequency is:
[tex]\omega = \sqrt{\frac{a_{max}}{A}}[/tex]
[tex]\omega = \sqrt{\frac{7\times 10^{3}\,\frac{m}{s^{2}} }{1.7\times 10^{-3}\,m} }[/tex]
[tex]\omega = 2.029\,\frac{rad}{s}[/tex]
The period of the motion is:
[tex]T = \frac{2\pi}{\omega}[/tex]
[tex]T = \frac{2\pi}{2.029\,\frac{rad}{s} }[/tex]
[tex]T = 3.097\,s[/tex]
b) The maximum speed of the particle is:
[tex]v_{max} = \omega \cdot A[/tex]
[tex]v_{max} = \left(2.029\,\frac{rad}{s}\right)\cdot (1.7\cdot 10^{-3}\,m)[/tex]
[tex]v_{max} = 3.449\times 10^{3}\,\frac{m}{s}[/tex]
c) The spring constant is:
[tex]k = \omega^{2}\cdot m[/tex]
[tex]k = (2.029\,\frac{rad}{s} )^{2}\cdot (0.092\,kg)[/tex]
[tex]k = 0.379\,\frac{N}{m}[/tex]
The total mechanical energy of the oscillator is:
[tex]E = \frac{1}{2}\cdot k \cdot A^{2}[/tex]
[tex]E = \frac{1}{2}\cdot \left(0.379\,\frac{N}{m}\right)\cdot (1.7\times 10^{-3}\,m)^{2}[/tex]
[tex]E = 5.477\times 10^{-7}\,J[/tex]
d) The magnitude of the force on the particle at its maximum displacement is:
[tex]F = k\cdot A[/tex]
[tex]F = (0.379\,\frac{N}{m})\cdot (1.7\times 10^{-3}\,m)[/tex]
[tex]F = 6.443\times 10^{-4}\,N[/tex]
e) The magnitude of the force on the particle at half of its maximum displacement is:
[tex]F = \frac{1}{2}\cdot k\cdot A[/tex]
[tex]F = (0.379\,\frac{N}{m})\cdot (0.85\times 10^{-3}\,m)[/tex]
[tex]F = 3.222\times 10^{-4}\,N[/tex]
