- #1
stunner5000pt
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Homework Statement
A mass hanging from a spring is displaced and released so that it vibrates vertically. Its
maximum height above a tabletop is 30 cm and its minimum height above the table top
is 12 cm. The mass vibrates 20 times per minute. At time 0, its height is 30 cm.
Calculate the maximum velocity of the mass, and the times at which it occurs
Homework Equations
s(t) represents displacement
v(t) represents s'(t) which is velocity
The Attempt at a Solution
Since we start at a maximum, we must use a cosine function
the period T is 1/20 min. [itex] k = 2\pi / T = 40\pi [/itex]
amplitude = (max - min) / 2 = [itex] \frac{30-12}{2} = 9 cm [/itex]
[tex] s(t) = 9\cos \left(40\pi t\right) + 21 [/tex]
[tex] v(t) = -360\pi \sin\left(40\pi t\right) [/tex]
[tex] a(t) = -1440 \pi^2 \cos\left(40\pi t\right) [/tex]
we find the max velocity when we solve a(t) = 0
so we need to solve
[tex] cos \left( 40\pi t \right) = 0 [/tex]
since cos is negative for all half pi values, and consider only one rotation, for now...
[tex] 40 \pi t = \frac{\pi}{2} \text{or} \frac{3\pi}{2}[/tex]
which yields [tex] t = \frac{1}{80} \text{or} \frac{3}{80} [/tex]
both answers have units of minutes.
Since the function v(t) first decreases and then increases, the critical point 3/80 is where the max occurs
if we plug each into v(t) we get
[tex] v \left(\frac{3}{80}\right) = 1131 cm/min
Is this all correct? Thank you for your help!