In simple harmonic motion, the total mechanical energy of the system is the sum of the potential energy and the kinetic energy. In this case, it is given that the kinetic energy at a distance of 2.5 cm from the equilibrium position is three-fourths of the total energy. Let's denote the total mechanical energy as E.
Since kinetic energy is three-fourths of the total energy, the potential energy at the same distance must be one-fourth of the total energy. This can be expressed as:
Potential Energy = (1/4) * E
In simple harmonic motion, the potential energy is directly proportional to the square of the displacement from the equilibrium position. Therefore, we can write:
Potential Energy = (1/4) * E = (1/2) * k * x^2
Where k is the spring constant and x is the displacement.
Now, let's consider the amplitude of the motion, denoted as A. The amplitude is the maximum displacement from the equilibrium position. In simple harmonic motion, the potential energy is maximum when the displacement is equal to the amplitude. So we have:
Potential Energy = (1/2) * k * A^2
Equating the expressions for potential energy, we can write:
(1/2) * k * A^2 = (1/4) * E
Simplifying the equation, we find:
k * A^2 = (1/2) * E
Now, let's consider the kinetic energy. The kinetic energy is also directly proportional to the square of the displacement. At a distance of 2.5 cm from the equilibrium position, the kinetic energy is given as three-fourths of the total energy, which can be expressed as:
Kinetic Energy = (3/4) * E = (1/2) * k * (2.5 cm)^2
Simplifying the equation, we have:
k * (2.5 cm)^2 = (3/2) * E
Now, we have two equations:
k * A^2 = (1/2) * E k * (2.5 cm)^2 = (3/2) * E
Dividing the second equation by the first equation, we get:
(2.5 cm)^2 / A^2 = (3/2) / (1/2)
Simplifying, we find:
A^2 = (2.5 cm)^2 * 2/3
Taking the square root of both sides, we obtain:
A = √[(2.5 cm)^2 * 2/3]
Calculating the value, we get:
A ≈ 2.04 cm
Therefore, the amplitude of the motion is approximately 2.04 cm.