Final answer:
The amplitude of the particle performing linear S.H.M is 25/(0.1π) cm and the period of oscillation is 2/(10π) seconds.
Step-by-step explanation:
To find the amplitude and period of the particle performing linear Simple Harmonic Motion (SHM), we can use the information provided about the maximum velocity and maximum acceleration.
The maximum velocity of 25 cm/s is given by the equation Vmax = A×ω, where A is the amplitude and ω is the angular frequency. Substituting the given values, we have 25 = A×0.1×π.
Simplifying the equation, we find that the amplitude A is 25/(0.1×π) cm.
The maximum acceleration of 100 cm/s² is given by the equation Amax = ω²A, where A is the amplitude and ω is the angular frequency. Substituting the given values, we have 100 = 0.1²×π²A.
From this equation, we can solve for the angular frequency ω, which is 10×π rad/s. Using the formula T = 2π/ω, we can find the period T of the oscillation, which is 2/(10×π) seconds.