- #1
Granger
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Homework Statement
The car circulates on a section of road whose profile can be approximated by a sinusoidal curve with the wavelength of 5.0 m. The mass of the car is 600.0 kg, and each wheel is equipped with a constant spring
k = 5000 Nm-1 and a damper with constant b = 450 Nm-1s.
Calculate the velocity of the car when the vertical oscillations have biggest amplitude.
Homework Equations
Equation of a damped oscillator
$$ x(t) = A e^{-\lambda t}\cos(\sqrt{\omega_0^2-\lambda^2}t +\phi)$$
where $$ \lambda=\frac{b}{2m} $$ and $$ \omega=\sqrt{\frac{k}{m}}$$
$$T=\frac{2\pi}{\omega_0}$$
The Attempt at a Solution
First thing I thought was that, since this is a damped oscillator, then the amplitude must be maximum at t=0.
Then since they give us the wavelength we know that x(0) = 0 and x(T) = 5.0 m
Substituting in the damped oscillator equation we get to:
$$ 0 = A\cos(\phi)$$
$$ 5= A e^{-\lambda T}\cos(\sqrt{\omega_0^2-\lambda^2}T - \phi)$$
Calculating all the known constants and solving this system of 2 equations we get to
$$\lambda=0.375$$
$$\omega_0 = 2.887$$
$$T=2.176$$
$$\phi=\pi/2$$
$$A=208.357$$
Now using the equation for velocity, by differentiating the equation of the damped oscillator:
$$ v(t) = -\lambdaA e^{-\lambda t}\cos(\sqrt{\omega_0^2-\lambda^2}t - \phi) - A \sqrt{\omega_0^2-\lambda^2} e^{-\lambda t}\cos(\sqrt{\omega_0^2-\lambda^2}t - \phi)$$
Substituting the know values we get to $$v=-596.43$$
The answer should be $$v=4.3$$
Can someone help me understand what am I doing wrong?
Thanks!