- #1
vxr
- 25
- 2
- 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?
##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?