Problem 2: A body A of mass m = 1Kg and a body B of
mass M = 4Kg are interconnected by a spring as shown in
figure below. The body A performs free vertical harmonic
oscillation with amplitude a = 1.6 cm and frequency ω=
25/s. Neglecting the mass of spring, find the maximum and
minimum values of force that this system exerts on the
bearing surface.
To find the maximum and minimum values of force exerted by the system on the bearing surface, we need to consider the forces acting on body A at different points in its motion.
1. At the extreme position: When body A is at the maximum displacement (maximum amplitude), the spring is stretched or compressed to its maximum length. At this point, the spring exerts the maximum force on body A in the opposite direction. This force is given by Hooke's Law: F = -kx, where F is the force exerted, k is the spring constant, and x is the displacement from the equilibrium position.
2. At the equilibrium position: When body A is at its equilibrium position (midpoint of the oscillation), the spring is neither stretched nor compressed, and there is no net force acting on body A due to the spring. Therefore, the force exerted by the spring at this point is zero.
3. At the extreme position in the opposite direction: When body A reaches the minimum displacement (opposite direction), the spring is again stretched or compressed to its maximum length, but in the opposite direction from before. At this point, the spring exerts the maximum force on body A in the opposite direction. This force is also given by Hooke's Law: F = -kx, where x is the displacement from the equilibrium position.
Since the mass of the spring is neglected, the force exerted by the spring is entirely due to the displacement of body A from its equilibrium position.
To find the spring constant, we can use the formula ω = √(k/m), where ω is the angular frequency and m is the mass of body A. Rearranging the formula, we get k = m * ω^2.
Given:
Mass of body A (m) = 1 kg
Angular frequency (ω) = 25/s
Substituting the values into the formula for k, we have:
k = 1 kg * (25/s)^2 = 625 N/m
Now we can find the maximum and minimum values of force exerted by the spring.
1. Maximum force:
At the extreme position, the displacement (x) is equal to the amplitude (a). Substituting the values, we have x = 1.6 cm = 0.016 m.
Using Hooke's Law, F = -kx, we can calculate the maximum force:
F_max = -(625 N/m) * (0.016 m) = -10 N
Note: The negative sign indicates that the force is in the opposite direction compared to the displacement.
2. Minimum force:
At the extreme position in the opposite direction, the displacement (x) is equal to the negative of the amplitude (-a). Substituting the values, we have x = -1.6 cm = -0.016 m.
Using Hooke's Law again, F = -kx, we can calculate the minimum force:
F_min = -(625 N/m) * (-0.016 m) = 10 N
Note: The negative sign cancels out, indicating that the force is in the same direction as the displacement.
Therefore, the maximum force exerted by the system on the bearing surface is 10 N (in the downward direction), and the minimum force exerted is -10 N (in the upward direction).