A 1-kg mass (the blue mass) is connected to a 9-kg mass (the green mass) by a massless rod 67 cm long, as shown in the figure. A hole is then drilled in the rod 40.2 cm away from the 1-kg mass, and the rod and masses are free to rotate about this pivot point, P. Calculate the period of oscillation for the masses if they are displaced slightly from the stable equilibrium position.
To calculate the period of oscillation for the masses, we can use the principle of simple harmonic motion (SHM). The period of a simple harmonic oscillator can be determined using the formula:
T = 2π √(m/k)
Where T is the period, m is the reduced mass, and k is the effective spring constant.
In this case, since the rod and masses are free to rotate about the pivot point P, we can treat it as a simple pendulum. The effective spring constant (k) can be calculated using the formula:
k = m * g * r
Where m is the reduced mass (m1 * m2 / (m1 + m2)), g is the acceleration due to gravity, and r is the distance between the center of mass and the pivot point P.
Let's calculate each component step-by-step:
Step 1: Calculate the reduced mass
The reduced mass (m) can be calculated by multiplying the masses of the two objects and dividing by their sum:
m = m1 * m2 / (m1 + m2)
In this case, m1 = 1 kg and m2 = 9 kg.
m = (1 kg * 9 kg) / (1 kg + 9 kg)
m = 0.9 kg
Step 2: Calculate the effective spring constant
The effective spring constant (k) can be calculated by multiplying the reduced mass, acceleration due to gravity, and the distance between the center of mass and the pivot point:
k = m * g * r
In this case, g is approximately 9.8 m/s², and r = 0.402 m (40.2 cm).
k = 0.9 kg * 9.8 m/s² * 0.402 m
k = 3.536 N/m
Step 3: Calculate the period of oscillation
Finally, we can use the calculated effective spring constant to determine the period of oscillation:
T = 2π √(m/k)
Substituting the values:
T = 2π √(0.9 kg / 3.536 N/m)
T ≈ 2π √(0.2551 s²)
T ≈ 2π * 0.505 s
T ≈ 3.183 s
Therefore, the period of oscillation for the masses is approximately 3.183 seconds.