Question 6.2: When at rest in equilibrium, the mass receives an impulse J ...
When at rest in equilibrium, the mass receives an impulse J of magnitude 5 Ns applied at time, t = 0. Find the response for t > 0. Data: k = 500 N/m c = 20 Ns/m m = 10 kg
Equation of motion
– 2kx – c\dot{x} = m\ddot{x}or m\ddot{x} + c\dot{x} + 2kx = 0
Note that the impulse J doesn’t appear on the free-body diagram or in the equation of motion since it doesn’t exist for t > 0.
From the coefficients in the equation of motion, we find that \omega _{n} = \sqrt{\frac{2k}{m} } = rad/s and \gamma = \frac{c} {2\sqrt{2km} } = 0.1 . Since the damping ratio is less than 1.0 in this problem, we can identify
that it is a case of light damping. From equation (6.42), we can write the expression for the response as:
z(t) = e^{- \gamma \omega _{n}t}(B_{1} \cos \omega_{n} \sqrt{1 – \gamma ^{2}} t + B_{2} \sin \omega_{n} \sqrt{1 – \gamma ^{2}} t ) (6.42)
x(t) = e^{- \gamma \omega _{n}t}(B_{1} \cos \Omega_{n} t + B_{2} \sin \Omega _{n} t)where \Omega _{n} = \omega _{n}\sqrt{1 – \gamma ^{2}} = 9.9 rad/s
First initial condition: x = 0 at t = 0 ∴ B_{1} = 0
Hence, x(t) = B_{2} e^{- \gamma \omega _{n}t} \sin \Omega _{n} t
The second initial condition is the velocity immediately after the impulse, \dot{x}_{0} . This can be found from the change in momentum caused by the impulse.
That is: J = m (\dot{x}_{0} – 0 )
Therefore, \dot{x}_{0} = \frac{J}{m}
Differentiating the expression for displacement gives
\dot{x} = B_{2} [\Omega_{n}e^{- \gamma \omega _{n}t} \cos \Omega _{n} t – \gamma \omega _{n} e^{- \gamma \omega _{n}t} \sin \Omega _{n} t]
\dot{x}_{0} = \frac{J}{m} at t = 0 ∴ \frac{J}{m} = B_{2}[\Omega _{n} – 0]
Hence, B_{2} = \frac{J}{m \Omega _{n}} and x(t) = \frac{J}{m \Omega _{n}} e^{- \gamma \omega _{n}t} \sin \Omega _{n} t
Substituting the numerical values gives x(t) = 0.0505 e^{-t} \sin 9.9t [m]