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## Homework Statement

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

## Homework Equations

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)

T=1/f

## The Attempt at a Solution

v(t)= -ωA sin(ωt+φ)

angular frequency=25.8199

amplitude=0.01m

time= 1/f=0.2433

phase constant= 0

v(t)= -25.8199 ×0.01 sin(25.8199 × 0.2433+0)

v(t)= -0.258199 sin(6.2817627+0)

v(t)= -0.02825ms^(-1)

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.**