Answer:
Step-by-step explanation:
The natural frequency of the oscillation of the rod-clamp system can be calculated using the formula:
f = 1/(2π) * sqrt(k/m)
where f is the natural frequency in hertz, k is the spring constant of the system, and m is the effective mass of the system.
In this case, the spring constant of the system can be approximated as the torsion constant of the aluminum rod. The torsion constant is given by:
k = (π/2) * G * r^4 / L
where G is the shear modulus of the aluminum, r is the radius of the rod, and L is the length of the rod.
Substituting the given values, we get:
k = (π/2) * 26.9 x 10^9 Pa * (1.25 cm)^4 / 30 cm = 0.637 Nm/rad
To determine the effective mass of the system, we need to consider the mass of the clamp and the mass of the portion of the rod that is oscillating. The effective length of the rod is the distance from the fixed end to the center of mass of the oscillating portion of the rod. Since the rod is uniform and symmetrical, the center of mass is at the midpoint of the oscillating portion, which is 15 cm from the fixed end. Therefore, the effective length of the rod is 15 cm.
The effective mass of the system can be calculated as:
m = m_clamp + m_rod
where m_clamp is the mass of the clamp and m_rod is the mass of the oscillating portion of the rod.
The mass of the clamp can be calculated as:
m_clamp = m_g / g = 20 kg / 9.81 m/s^2 = 2.04 kg
The mass of the oscillating portion of the rod can be calculated as the mass of a cylinder with radius 1.25 cm and length 15 cm:
m_rod = π * (1.25 cm)^2 * 15 cm * 2.7 g/cm^3 = 2.53 kg
Therefore, the effective mass of the system is:
m = m_clamp + m_rod = 2.04 kg + 2.53 kg = 4.57 kg
Substituting the values of k and m into the formula for the natural frequency, we get:
f = 1/(2π) * sqrt(k/m) = 1/(2π) * sqrt(0.637 Nm/rad / 4.57 kg) = 0.259 Hz
Therefore, the frequency of the oscillations of the rod-clamp system is 0.259 Hz.