Answers:
a) [tex]0.80 kg[/tex]
b) [tex]2.12 s[/tex]
c) [tex]1.093 m/s^{2}[/tex]
Explanation:
We have the following data:
[tex]k=7 N/m[/tex] is the spring constant
[tex]A=12.5 cm \frac{1 m}{100 cm}=0.125 m[/tex] is the amplitude of oscillation
[tex]V=32 cm/s=0.32 m/s[/tex] is the velocity of the block when [tex]x=\frac{A}{2}=0.0625 m[/tex]
Now let's begin with the answers:
We can solve this by the conservation of energy principle:
[tex]U_{o}+K_{o}=U_{f}+K_{f}[/tex] (1)
Where:
[tex]U_{o}=k\frac{A^{2}}{2}[/tex] is the initial potential energy
[tex]K_{o}=0[/tex] is the initial kinetic energy
[tex]U_{f}=k\frac{x^{2}}{2}[/tex] is the final potential energy
[tex]K_{f}=\frac{1}{2} m V^{2}[/tex] is the final kinetic energy
Then:
[tex]k\frac{A^{2}}{2}=k\frac{x^{2}}{2}+\frac{1}{2} m V^{2}[/tex] (2)
Isolating [tex]m[/tex]:
[tex]m=\frac{k(A^{2}-x^{2})}{V^{2}}[/tex] (3)
[tex]m=\frac{7 N/m((0.125 m)^{2}-(0.0625 m)^{2})}{(0.32 m/s)^{2}}[/tex] (4)
[tex]m=0.80 kg[/tex] (5)
The period [tex]T[/tex] is given by:
[tex]T=2 \pi \sqrt{\frac{m}{k}}[/tex] (6)
Substituting (5) in (6):
[tex]T=2 \pi \sqrt{\frac{0.80 kg}{7 N/m}}[/tex] (7)
[tex]T=2.12 s[/tex] (8)
The maximum acceleration [tex]a_{max}[/tex] is when the force is maximum [tex]F_{max}[/tex], as well :
[tex]F_{max}=m.a_{max}=k.x_{max}[/tex] (9)
Being [tex]x_{max}=A[/tex]
Hence:
[tex]m.a_{max}=kA[/tex] (10)
Finding [tex]a_{max}[/tex]:
[tex]a_{max}=\frac{kA}{m}[/tex] (11)
[tex]a_{max}=\frac{(7 N/m)(0.125 m)}{0.80 kg}[/tex] (12)
Finally:
[tex]a_{max}=1.093 m/s^{2}[/tex]
