Mathematics: An elastic band is hung on a hook and a mass is hung on the lower end of the band. When the mass is pulled down and then released, it vibrates vertically. The equation of motion, s, where s is measured in centimeters and t≥0 in seconds, is given below. (Take the positive direction to be downward.)s = 8 cos (t) + 4 sin (t)I found velocity and accelerationv(t)= -8sin(t) +4cos(t)a(t)= -8cos(t) - 4sin(t)MY problem comes in the question c(c) When does the mass pass through the equilibrium position for

q = 8Step-by-step explanation:...

An elastic band is hung on a hook and a mass is

Ответ:

gigi813

19.01.2022

Mathematics

Step-by-step explanation:

Velocity and acceleration

s = 8cos(t) + 4sin(t)

v = -8sin(t) + 4cos(t)

a = -8cos(t) + 4sin(t)

The equilibrium position is where the mass hangs before it is pulled downward, that is, at s = 0.

Set s = 0 and solve for t.

$\begin{array}{rcl}0 & = & 8\cos t + 4\sin t\\-8\cos t & = & 4\sin t\\-8 & = & 4\dfrac{\sin t}{\cos t}\\\\-2 & = & \tan t\\t & = & \arctan(-2)\\& = & -1.107 \pm n\pi\\\end{array}$

If n = 1,

t = -1.107 + π = 2.034 s

(d) Distance from equilibrium position

The mass will reach its maximum distance when v = 0, that is, when it is at the peak or trough of its oscillation.

Set v = 0 and solve for t.

$\end{array}\begin{array}{rcl}0 & = &- 8\sin t + 4\cos t\\8\sin t & = & 4\cos t\\\dfrac{\sin t}{\cos t} & = & \dfrac{1}{2}\\\\\tan t & = & 0.5\\t & = & \arctan(0.5)\\& = & 0.4636 \pm n\pi\\\end{array}$

If n = 0,

t = 0.4636

Then

s = 8cos(0.4636) + 4sin(0.4636) = 8×0.8944 + 4×0.4472 = 7.156 + 1.789 = 8.944 cm

The figure below shows the graphs of s and v vs t. They indicate that the mass first reaches its equilibrium position at 2.034 s, and the amplitude of its vibration is 8.944 cm.

An elastic band is hung on a hook and a mass is hung on the lower end of the band. When the mass is

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