Question

g 1. A mass undergoing simple harmonic motion along the x-axis has a period of T = 0.5 s and an amplitude of 25 mm. Its position is x = 14 mm when t = 0. a. Write down x(t) = A cos(ωt + φ0) for this oscillator, filling in A, ω, and φ0. b. What is the magnitude of the maximum velocity vmax? At what value(s) of x does it occur? c. What is the magnitude of the maximum acceleration amax? At what value(s) of x does it occur?

Answer 1

Answer:

Explanation:

a )

Amplitude A = 14 mm , angular frequency ω = 2π / T

= 2π / .5

ω = 4π rad /s

φ₀ = initial phase

Putting the given values in the equation

14 = 25 cos(ωt + φ₀ )

14/25 = cosφ₀

φ₀ = 56 degree

x(t) = 25cos(4πt + 56° )

b )

maximum velocity = ω A

=  4π  x 25

100 x 3.14 mm /s

= 314 mm /s

At x = 0 ( equilibrium position or middle point , this velocity is achieved. )

maximun acceleration = ω² A

= 16π² x A

= 16 x 3.14² x 25

= 3943.84 mm / s²

3.9 m / s²

It occurs at x = A or at extreme position.