An object of mass 5.00 kg is attached to a spring with a force constant of 300 N/m. It is oscillating on a horizontal frictionless surface with an amplitude of 2.00 m. A 8.00 kg object is dropped vertically on top of the 5.00 kg object as it passes through its equilibrium point. The two objects stick together. By how much does the amplitude of the vibrating system change?
Why would the amplitude change? The deflection is in the horizontal. Adding mass should not change that deflection (but will change the period (frequency)).
To find out how much the amplitude of the vibrating system changes, we need to analyze the conservation of energy before and after the collision between the two objects.
Before the collision, the 5.00 kg object is oscillating with an amplitude of 2.00 m. The total initial mechanical energy of the system is the sum of potential energy and kinetic energy:
E_initial = 1/2 * k * A^2 + 1/2 * m * v^2
where k is the force constant of the spring, A is the amplitude of the oscillation, and v is the velocity of the 5.00 kg object.
After the collision, when the 8.00 kg object is dropped onto the 5.00 kg object, they stick together and move as a single mass. Let's call this mass M, equal to the sum of the masses:
M = m1 + m2
where m1 is the initial mass of the 5.00 kg object and m2 is the mass of the 8.00 kg object.
Now, we need to find the new amplitude, A_new, of the vibrating system.
The total final mechanical energy, E_final, of the vibrating system after the collision is given by:
E_final = 1/2 * k * A_new^2 + 1/2 * M * v_final^2
where v_final is the velocity of the vibrating system after the collision.
According to the conservation of energy, the initial mechanical energy equals the final mechanical energy:
E_initial = E_final
Therefore, we can write:
1/2 * k * A^2 + 1/2 * m * v^2 = 1/2 * k * A_new^2 + 1/2 * M * v_final^2
Simplifying the equation, we have:
1/2 * k * A^2 = 1/2 * k * A_new^2 + 1/2 * M * v_final^2
Now, let's consider the motion at the equilibrium point. At the equilibrium point, the velocity of the 5.00 kg object is zero, as it momentarily stops. Therefore, the velocity of the 8.00 kg object just before impact is equal to the velocity of the 5.00 kg object:
v_final = v
Plugging this into the equation, we get:
1/2 * k * A^2 = 1/2 * k * A_new^2 + 1/2 * M * v^2
Since the two objects stick together, their combined mass M = m1 + m2 is equal to 5.00 kg + 8.00 kg = 13.00 kg.
Now, let's rearrange the equation to solve for A_new:
1/2 * k * A^2 - 1/2 * k * A_new^2 = 1/2 * M * v^2
Multiplying each term by 2 and dividing by k, we have:
A^2 - A_new^2 = M * v^2 / k
Substituting the given values, A = 2.00 m, M = 13.00 kg, v = 0 (since the 5.00 kg object momentarily stops at the equilibrium point), and k = 300 N/m, we can now solve for A_new:
2.00^2 - A_new^2 = 13.00 * 0^2 / 300
4.00 - A_new^2 = 0
A_new^2 = 4.00
Taking the square root of both sides to find the positive value of A_new, we get:
A_new = 2.00 m
Therefore, the amplitude of the vibrating system remains the same at 2.00 m even after the collision and the two objects stick together.