I am new to vibration analysis/simulation etc. so I may be missing something very basic here so please forgive me if I'm explaining things poorly or not making sense.

To familiarize myself with vibration simulation I am trying to simulate the frequency response spectrum for a tuning fork. I am using Solidworks. I have

  • created the geometry for a tuning fork
  • fixed the base and calculated the eigenfrequencies eigenmode shapes
  • Applied a uniform base displacement excitation (same magnitude for all frequencies)
  • Plotted the lateral displacement of one of the tines as a function of excitation frequency
  • The part is made from copper (young modulus supplied by software) and I am using a modal damping ratio of 0.02 for all modes.
  • The excitation amplitude is 1mm

Here is an image of the main "tuning fork" mode showing the fixed base (green arrows) and excitation vector (purple arrow).enter image description here

Here is a plot of the magnitude lateral displacement (x direction) of a node at the tip of one of tines of the tuning force versus excitation frequency.

enter image description here

The peak at ~500 Hz corresponds to the mode shown above. The peak at ~2.5 kHz corresponds to a higher frequency vibrational mode.


I am confused about the high frequency response of the tuning fork. At low frequency I see that the response is falling off from the two resonances towards lower frequencies. This makes sense. If you slowly move the tuning fork up and down (from the base) there will be no stress or deflection since everything is slow so there will be no lateral displacement.

However, I would ALSO expect the response to drop off at high frequencies as well.. I guess my intuition here is motivated by the following. Consider a mass attached to a base by a spring and a damper. If the base is shaken then at high frequencies the response of the mass falls off, it is as if there is nothing connecting the mass and the base at high frequency. I guess I expect something similar for the tuning fork. If I drive at high enough frequency the tuning forks can't even tell there is a drive..

Can someone please explain to me:

1) If it is correct that this response function should be constant versus frequency above the tuning fork resonance frequencies and

2) Whether it is correct or incorrect can you give me any intuition for why we should expect whatever is the correct behavior?

  • $\begingroup$ I don't have the answer for you, but here's how I would go about investigating: plot the displaced shape at very high frequency. i.e. not the eigenmode, but the "operating deflection shape". This may give you a clue that you aren't getting by merely looking at a single DOF of one part of the structure. $\endgroup$
    – Daniel K
    Commented Nov 27, 2018 at 2:09

1 Answer 1


You said you applied a constant displacement to the base. In that situation, if the structure doesn't have any modes that can are excited at a particular frequency, the response is going to be approximately the same as a rigid body being moved with a fixed amplitude, which is what your plot shows.

Note that if you apply a fixed displacement, the excitation force will increase for higher frequencies (and except where there is a resonance in the displacement response) it will be approximately proportional to the frequency squared.

If you applied a constant force, you would see the displacement falling as the frequency increased, but unless the structure is constrained in some way, the displacement amplitude will increase to infinity as the frequency gets close to zero.

Another way to understand this is think abut a 2 DOF system being excited at a frequency much higher than its two resonant frequencies. In that situation the elastic forces are small compared with the inertia forces, so we can ignore the stiffness term in the equation of motion and approximate it as $$\begin{bmatrix} m_{11} & m_{12} \\ m_{21} & m_{22} \end{bmatrix} \begin{bmatrix} \ddot x_1 \\ \ddot x_2 \end{bmatrix} = \begin{bmatrix}F \\ 0 \end{bmatrix} $$ where F is the excitation force applied at point 1.

Now if we apply a constant amplitude displacement $X$ at point 1, we have $\ddot x_1 = -\omega^2X$ and the two unknowns in the equations are $\ddot x_2$ and $F$.

From the second equation we have $$\ddot x_2 = - \frac{m_{21}}{m_{22}} \ddot x_1$$ which means that $$x_2 = - \frac{m_{21}}{m_{22}}X.$$

In other words, for a constant base amplitude, the tip amplitude of your tuning fork will also be approximately constant at high frequencies, except when there is a resonance that affects the motion of the system.

  • $\begingroup$ I understand your last two paragraphs. I also almost understand your first paragraph. I agree that if we are driving above the natural resonances of the structure that we expect a rigid body being moved with a fixed amplitude. In this case, since the force I am applying is in the Y direction I would expect the whole body to move up and down as a rigid body with motion purely in the Y direction.. However, the plot is showing the X-displacement of one of the tines is non-zero and constant at high frequency. This breaks my intuition that there should be only rigid body motion in the Y direction. $\endgroup$
    – Jagerber48
    Commented Nov 27, 2018 at 1:19
  • $\begingroup$ Explaining this is too long for a comment, so I edited my answer. $\endgroup$
    – alephzero
    Commented Nov 27, 2018 at 12:47
  • $\begingroup$ Hmm ok so it seems like the answer relies on the off-diagonal elements of the mass matrix, $m_{21}$. The simple models I have been considering (one dimensional stacks of oscillators) don't capture the essential features. A simpler model which also shows this behavior is a single oscillator attached by a spring to a base. If the base is shaken transverse to the connection the mass will oscillate transverse to the connection as well. If one then asks what is the VERTICAL motion of the base one will see that there is again flat transfer at high frequency. $\endgroup$
    – Jagerber48
    Commented Nov 29, 2018 at 7:55
  • $\begingroup$ I guess this is directly related to what you have shown here. $\endgroup$
    – Jagerber48
    Commented Nov 29, 2018 at 7:56

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