Answer:
f = 3.09 Hz
Explanation:
This is a simple harmonic motion exercise where the angular velocity is
w² = [tex]\frac{k}{m}[/tex]
to find the constant (k) of the spring, we use Hooke's law with the initial data
F = - kx
where the force is the weight of the body that is hanging
F = W = m g
we substitute
m g = - k x
k = [tex]- \frac{m g}{x}[/tex]
we calculate
k = [tex]- \frac{9.8 m}{- 2.6 \ 10^{-2}}[/tex]
k = 3.769 10² m
we substitute in the first equation
w² = [tex]\frac{ 3.769 \ 10^2 \ m }{m}[/tex]
w = 19.415 rad / s
angular velocity and frequency are related
w = 2πf
f = [tex]\frac{w}{2\pi }[/tex]
f = 19.415 / 2pi
f = 3.09 Hz
