Final answer:
To answer the student's questions, use conservation of energy for part a), the formula involving spring constant and mass for part b), a simple potential energy calculation for part c), and acknowledge that the amplitude found in part a) is the maximum displacement in part d).
Step-by-step explanation:
The question requires finding various properties of an object undergoing simple harmonic motion (SHM) attached to a spring. The object has a mass of 2.9 kg and is attached to a spring with a spring constant of 270 N/m. When the object is 0.025 m from its equilibrium position, it has a speed of 0.45 m/s.
a) Find the amplitude of motion
To find the amplitude of motion, we use conservation of energy. The total mechanical energy (E) is constant in SHM, being the sum of kinetic (K) and potential energy (U). We know that U = 1/2 k x2 and K = 1/2 m v2, where k is the spring constant, x is the displacement, m is the mass, and v is the velocity. Therefore, E = U + K. At the point of maximum displacement (the amplitude, A), all energy is potential, so E = 1/2 k A2 = 1/2 k x2 + 1/2 m v2. Solving this equation for A when x=0.025 m and v=0.45 m/s gives A = sqrt(x2 + (m v2/k)).
b) Calculate the frequency of the oscillation
The frequency (f) can be found using the formula f = 1/(2π) √(k/m), where k is the spring constant and m is the mass of the object. We would plug in the given values for k and m to find f.
c) Determine the potential energy of the system at the given position
The potential energy (U) at any displacement x from the equilibrium position is given by U = 1/2 k x2. We substitute x = 0.025 m and k = 270 N/m to find U.
d) Find the maximum displacement from the equilibrium position
The maximum displacement from the equilibrium position is the amplitude (A), which is found in part a).