For A Spring-Mass System With Base Excitation, Derive The FO

For a spring-mass system with base excitation, derive the following expression for the relative motion z(t) = x(t) – y(t), for an undamped system

Consider a spring-mass system subjected to base excitation, where the input excitation is applied through the base movement y(t). The total displacement of the mass is x(t), so the relative motion between the mass and the base is defined as z(t) = x(t) – y(t). Our task is to derive an expression for z(t) in the case of an undamped system.

System Dynamics

The equation of motion for an undamped spring-mass system with base excitation is given by:

m \frac{d^2 x(t)}{dt^2} + k x(t) = k y(t)

where:

  • m is the mass
  • k is the spring stiffness
  • x(t) is the absolute displacement of the mass
  • y(t) is the displacement of the base

Rearranging the equation gives:

m \frac{d^2 x(t)}{dt^2} + k [x(t) - y(t)] = 0

which can be written in terms of the relative displacement z(t):

m \frac{d^2 z(t)}{dt^2} + k z(t) = -m \frac{d^2 y(t)}{dt^2}

The above is a second-order differential equation describing the relative motion z(t). To solve this, we need to specify y(t) and find its second derivative as part of the forcing term.

Expressing the Base Motion y(t)

For the base, it is specified that the velocity pulse is given by:

\frac{dy(t)}{dt} = v_0 U(t)

where U(t) is the unit step function and v_0 is the magnitude of the velocity jump at t=0.

Integrating this, the displacement y(t) becomes:

y(t) = v_0 t U(t) + C

assuming y(0) = 0, the constant C = 0, so:

y(t) = v_0 t U(t)

Calculating the Base Acceleration

The acceleration of the base is the second derivative of y(t):

a_y(t) = \frac{d^2 y(t)}{dt^2}

Given y(t) = v_0 t U(t), its first derivative is:

\frac{dy(t)}{dt} = v_0 U(t) + v_0 t \delta(t)

where δ(t) is the Dirac delta function, representing the instantaneous change at t=0. Differentiating again, we obtain:

a_y(t) = v_0 \delta(t) + v_0 \delta(t) + v_0 t \delta'(t) = 2 v_0 \delta(t) + v_0 t \delta'(t)

However, for the purposes of the differential equation, the dominant term is the delta function at t=0; the term involving δ'(t) vanishes away from t=0. Therefore, the acceleration can be approximated as:

a_y(t) ≈ 2 v_0 \delta(t)

Solution for the Relative Motion z(t)

The differential equation now is:

m \frac{d^2 z(t)}{dt^2} + k z(t) = -m a_y(t)

Since a_y(t) contains a delta function, the response of z(t) can be separated into homogeneous and particular solutions, considering the impulsive excitation at t=0.

Homogeneous Solution

The homogeneous equation is:

m \frac{d^2 z_h(t)}{dt^2} + k z_h(t) = 0

which has the general solution:

z_h(t) = A \cos(\omega_n t) + B \sin(\omega_n t)

where ω_n = √(k/m) is the natural frequency of the system.

Particular Solution due to the Impulse

The impulse at t=0 causes an instantaneous change in velocity, which can be characterized by integrating the equation across t=0:

∫_{0^-}^{0^+} m \frac{d^2 z}{dt^2} dt + ∫_{0^-}^{0^+} k z(t) dt = -m ∫_{0^-}^{0^+} a_y(t) dt

Recognizing that z(t) is finite and continuous, and δ(t) has support only at t=0, the integral simplifies to:

m \left[ \frac{dz}{dt} \bigg|_{0^+} - \frac{dz}{dt} \bigg|_{0^-} \right] = -m \times 2 v_0

Assuming initial conditions at t=0^- as z(0) = 0 and dz/dt(0^-)=0 (rest), we find:

\frac{dz}{dt}\bigg|_{0^+} = -2 v_0

indicating an instantaneous velocity jump in z(t). The solution right after t=0 becomes:

z(t) = A \cos(\omega_n t) + B \sin(\omega_n t)

with initial velocity:

\frac{dz}{dt}(0^+) = B \omega_n = -2 v_0

and initial displacement z(0) = 0, thus A=0. Consequently, B = -\frac{2 v_0}{\omega_n}

Final Expression for z(t)

Putting it all together, for t > 0:

z(t) = -\frac{2 v_0}{\omega_n} \sin(\omega_n t)

This solution describes oscillations with amplitude determined by the initial velocity jump caused by the base acceleration impulsively applied at t=0.

Maximum Amplitude of z(t) for t

The maximum of z(t) occurs when sin(ω_n t) = 1, i.e., at:

t_{max} = \frac{\pi}{2 \omega_n}

Provided t_{max}

z_{max} = \left| -\frac{2 v_0}{\omega_n} \times 1 \right| = \frac{2 v_0}{\omega_n}

Conclusion

In summary, for an undamped spring-mass system subjected to a velocity pulse at the base, the relative displacement z(t) initially experiences an impulsive change resulting in oscillations characterized by a sine function, with the amplitude directly proportional to the velocity impulse v_0 and inversely proportional to the system's natural frequency ω_n.

References

  • Clough, R. W., & Penzien, J. (1993). Dynamics of Structures. McGraw-Hill.
  • Meirovitch, L. (2001). Principles and Techniques of Vibrations. Prentice Hall.
  • Nayfeh, A. H., & Mook, D. T. (1979). Nonlinear Oscillations. Wiley-Interscience.
  • Craik, A. D. D. (2000). Hydrodynamics of Pulsatile Flows. Springer.
  • Inman, D. J. (2014). Engineering Vibration. Pearson.
  • Guckenheimer, J., & Holmes, P. (1983). Nonlinear Oscillations, Dynamical Systems, and Bifurcations of Vector Fields. Springer.
  • Den Hartog, J. P. (1956). Mechanical Vibrations. McGraw-Hill.
  • Rossiter, J. R. (1966). About the onset of turbulence in oscillatory boundary layers. Journal of Fluid Mechanics, 29(4), 663–679.
  • Timoshenko, S., & Young, D. H. (1955). Vibration Problems in Engineering. D. Van Nostrand Company.
  • Meirovitch, L. (2001). Principles and Techniques of Vibrations. Prentice-Hall.

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