Final answer:
The frequency of damped oscillations in the horizontal spring-mass system is approximately 3.06 Hz. The number of cycles corresponding to a 50% reduction in amplitude is approximately 0.227 seconds.
Step-by-step explanation:
i. To find the frequency of damped oscillations in a horizontal spring-mass system with Coulomb damping, we can use the formula:
f = (√(k/m)^2 - (2πξ)^2)
Where f is the frequency, k is the spring stiffness, m is the mass, and ξ is the coefficient of friction. Plugging in the values given, we have:
f = (√((980 N/m) / 5 kg)^2 - (2π * 0.25)^2)
Simplifying, we find that the frequency of damped oscillations is approximately 3.06 Hz.
ii. To calculate the number of cycles corresponding to a 50% reduction in amplitude, we can use the formula:
n = t * f
Where n is the number of cycles, t is the time period, and f is the frequency. Since the frequency is given and we need to find the time period, we can rearrange the formula:
t = n / f
Given that the initial amplitude is 5 cm and the 50% reduction corresponds to half of the initial amplitude, we have:
n = log(0.5) / log(0.5)
Substituting the values into the formula, we get:
t = (log(0.5) / log(0.5)) / 3.06 Hz
Simplifying, we find that the time taken to achieve a 50% reduction in amplitude is approximately 0.227 seconds.