Final answer:
The amplitude of the oscillator's motion, given the mass, velocity, and spring constant, is calculated using the relationship between kinetic and potential energy in simple harmonic motion. The amplitude is found to be 0.49 m.
Step-by-step explanation:
The question addresses the amplitude of an oscillator's motion. In simple harmonic motion, the kinetic energy at the equilibrium position is converted entirely to potential energy at the maximum displacement, i.e., the amplitude.
The total mechanical energy E in a spring-mass system can be given by the formula E = 1/2*k*A^2, where k is the spring constant and A is the amplitude. At the equilibrium position, the mechanical energy is solely kinetic, E = 1/2*m*v^2, where m is the mass and v is the velocity. Setting these two energies equal to each other gives 1/2*k*A^2 = 1/2*m*v^2, which simplifies to A = v*sqrt(m/k).
Plugging in the given values, we have:
- m = 2.2 kg (Mass)
- v = 1.1 m/s (Velocity through equilibrium)
- k = 11 N/m (Spring constant)
Amplitude (A) = 1.1 m/s * sqrt(2.2 kg / 11 N/m) = 1.1 m/s * sqrt(0.2 kg/Nm) = 1.1 m/s * 0.4472 s = 0.49 m. So, the amplitude of this oscillator's motion is 0.49 m.