A mass of 0.3kg is suspended from a spring of stiffness 200Nm-1. If the mass is displaced by 10mm from its equilibrium position and released, for resulting vibration, calculate:
a) the maximum velocity of the mass during the vibration
Angular frequency= √((Spring constant)/mass)
Frequency= (Angular frequency)/(2 × π)
Velocity: v(t) = -ωA sin(ωt + φ)
Acceleration: a(t) = -ω 2 A cos(ωt + φ)
Displacement = amplitude x sin(angular frequency x time)
The Attempt at a Solution
v(t)= -ωA sin(ωt+φ)
phase constant= 0
v(t)= -25.8199 ×0.01 sin(25.8199 × 0.2433+0)
v(t)= -0.258199 sin(6.2817627+0)
Now i seem to have confused myself and could really do with some help to point me in the right direction.
I need to calculate what the maximum velocity would be. I've read that Maximum and minimum values of any sine and cosine function are +1 and -1. So, would the correct equation be v(t)= -0.258199 x 0.01 sin(1)? Furthermore I've read velocity is at a maximum when displacement is zero. By my understanding this would be when the amplitude is at 0.00. In which case should the equation be v(t)= -25.8199 ×0.00 sin(25.8199 × 0.2433+0).
Both ways seem to give a velocity of 0.00ms?
And is there any other factors that then need adjusting if amplitude is changed? doesn't seem to be, but thought i'd check.
If anybody is able to offer their help that would be great. Thanks.