A mass of 1.25 kg stretches a spring 0.05 m. The mass is in a medium that exerts

Question

A mass of 1.25 kg stretches a spring 0.05 m. The mass is in a medium that exerts a viscous resistance of 15 N when the mass has a velocity of 6 m/s. The viscous resistance is proportional to the speed of the object.

Suppose the object is displaced an additional 0.04 m and released.

Find an function to express the object's displacement from the spring's equilibrium position, in m after t seconds. Let positive displacements indicate a stretched spring, and use 9.8 m/s^2 as the acceleration due to gravity.

PLZ LIST THE DETAIL STEPS OF YOUR SOLUTION

I WILL RATE 5 STARS IF THE ANSWER IS RIGHT

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Explanation / Answer

we have force = -15 = p*6
p = -2.5
so force due to viscous forces = p* v = -2.5 * v
we have F = -kx - (2.5*v)
we have mg = kx0
1.25 * 9.8 = k* 0.05
k=245
so, m * (d^2 x/ dt^2) = (-245 * x)- (2.5 * (dx/dt))
1.25 * (d^2 x/ dt^2) = (-245 * x)- (2.5 * (dx/dt))
we have initial position, x = 0.04 m ; initial velocity = dx/dt = 0
x = c1 * (e^-t) * sin (13.9642 t) + c2* (e^-t) * cos (13.9642 t)
we have x = 0.04 m at t =0
0.04 = c2
x = c1 * (e^-t) * sin (13.9642 t) + 0.04* (e^-t) * cos (13.9642 t)
dx/dt = c1 * (e^-t) * 13.9642 * cos (13.9642 t) - c1 * (e^-t) * sin (13.9642 t) - 0.04 * (e^-t) * 13.9642 * sin (13.9642 t) - 0.04 * (e^-t) * cos (13.9642 t)
so at t =0 ,
dx/dt = 13.9642 * c1 - 0.04 = 0
c1 = 0.0028644677
x = (0.0028644677 * (e^-t) * sin (13.9642 t)) + (0.04* (e^-t) * cos (13.9642 t))
u and x represent same which is displacement from the equilibrium position.
u (t) = (0.0028644677 * (e^-t) * sin (13.9642 t)) + (0.04* (e^-t) * cos (13.9642 t))

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