Final answer:
The amplitude of an oscillating mass described by the function x(t) = (2.5 cm)cos(13t) is 2.5 cm. This figure represents the maximum displacement from the equilibrium position during its simple harmonic motion.
Step-by-step explanation:
The question presented pertains to the physical concept of oscillation, specifically relating to a 50 g (0.05 kg) mass executing simple harmonic motion (SHM). The position of the oscillating mass is given by a cosine function x(t) = (2.5 cm)cos(13t), where 'x' represents the displacement from the equilibrium position at a given time 't'. The amplitude, 'A', of this oscillatory motion is an essential characteristic of SHM and corresponds to the maximum displacement of the mass from its equilibrium position. In the provided equation, this value can be found directly as the coefficient of the cosine function.
The amplitude of the given oscillatory motion is 2.5 cm. This parameter dictates how far the mass travels from the center point of the oscillation during its motion. The student has been provided with an additional information correlating to the energy of the particle when it is at the equilibrium position. Using a modified version of the kinetic energy formula, we observe that the energy of the particle at the center (E) is calculated to be 2.5 J (joules), using the mass of the particle and its velocity squared.
The position versus time graph provided likely illustrates how the cosine function models the position of the mass within a decaying oscillation, possibly indicating a damped harmonic motion where an exponential term would describe the decrease in amplitude over time.