Mechanical vibrations: maximum velocity

[vxr]

Homework Statement

An air-track glider is attached to a spring pulled at distance $d = 0.2m$ to the right. Starting from released stage at $t = 0$ it subsequently makes $n = 15$ complete oscillations in time $t = 10s$. Determine the period of oscillation $T$ and the object's maximum speed (velocity) $v$, as well as its position and velocity $v$ at time $t = 0.8s$.

Relevant Equations

$x(t) = Acos(\omega t + \theta)$

So I am almost sure I know how to solve this, just curious about the maximum velocity. Anyway, if you could double check my calculations, here it is.

$T = \frac{t}{n} = \frac{10s}{15} = \frac{2}{3}s$

$\omega = \frac{2\pi}{T} = 2\pi \frac{3}{2} = 3\pi$

a). position at $t = 0.8s$:

$x(t) = Acos{(\omega t + \theta)} = Acos\Big(\frac{2\pi t}{T} + \theta\Big) = Acos\Big( 3\pi t \Big)$

$x(0.8s) = 0.2 cos (7,54) = 0.061 m$

b). velocity at $t = 0.8s$:

$v = \frac{d}{dt}x = \frac{d}{dt}\Big( Acos(\omega t + \theta) \Big) = -A\omega sin(\omega t + \theta)$

$v(0.8s) = -0.2 * 3\pi sin (3\pi * 0.8) = -\frac{3}{5} sin (\frac{12}{5}\pi) =~ -1.79 \frac{m}{s}$

c). maximum velocity. During my classes something like that was done:

$v_{max} = A\omega \cos (0) = A \omega = 0.2 * 9.42 = 1.884 \frac{m}{s}$

I understand that $cos(0) = 1$ but is this correct?

 

[.Scott]

Yes. You are correct.
You have $v= -A\omega sin(f(t))$
The largest this value can be is when the sin is -1.
Hence: $A\omega$

 

[vxr]

Thanks.