Answer:
[tex]w=\sqrt{\frac{k}{m} }[/tex]
b. [tex]5.6rad/s[/tex]
Explanation:
a. from the spring-mass system which is explicitly describe by hooks law
from
F=-kx
which is in comparison to newtons general law of motion
F=ma
where the displacement x is expressed as
[tex]x=Asin(wt)\\[/tex]
and the acceleration is the second derivative of the displacement
[tex]a=-Aw^{2}sin(wt)\\[/tex]
hence final expression after substituting for the acceleration and the displacement is expressed as
[tex]w=\sqrt{\frac{k}{m} }[/tex]
b. for k=105N/m and m=3.4kg
we have the angular frequency to be
[tex]w=\sqrt{\frac{105}{3.4}}\\\\w=5.6rad/s[/tex]
