Answer:
255.4 N/m
Explanation:
We can consider the system eyeball-attached to the musculature as a mass-spring system in simple harmonic motion, whose frequency of oscillation is given by
[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]
where in this case, we know:
f = 29 Hz is the frequency of oscillation
k is the spring constant, which is unknown
m = 7.7 g = 0.0077 kg is the mass of the eyeball
Solving the equation for k, we find the spring constant of the musculature attached to the eyeball:
[tex]k=(2\pi f)^2 m=(2 \pi (29 Hz))^2 (0.0077 kg)=255.4 N/m[/tex]
