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3.3. Mechanical elements

Mechanical circuit elements need not always be represented by electrical symbols. Because one frequently draws a mechanical circuit directly from inspection of the mechanical device, more obvious forms of mechanical elements are sometimes useful, at least until the student is thoroughly familiar with the analogous circuit. We shall accordingly devise a set of “mechanical” elements to be used as an introduction to the elements of Table 3.1.

In electrical circuits, a voltage measurement is made by attaching the leads from a voltmeter across the two terminals of the element. Voltage is a quantity that we can measure without breaking into the circuit. To measure electric current, however, we must break into the circuit because this quantity acts through the element. In mechanical devices, on the other hand, we can measure the velocity (or the displacement) without disturbing the machine by using a capacitive or inertially operated vibration pickup to determine the quantity at any point on the machine. It is not velocity but force that is analogous to electric current. Force cannot be measured unless one breaks into the device.

It becomes apparent then that if a mechanical element is strictly analogous to an electrical element, it must have a velocity difference appearing between (or across) its two terminals and a force acting through it. Analogously, also, the product of the rms force f in N and the in-phase component of the rms velocity u in m/s is the power in W. We shall call this type of analogy, in which a velocity corresponds to a voltage and a force to a current, the admittance-type analogy. It is also known as the “inverse” analogy.

Many texts teach in addition a “direct” analogy. It is the opposite of the admittance analogy in that force is made to correspond to voltage and velocity to current. In this chapter, we shall call this kind of analogy an impedance-type analogy. To familiarize the student with both concepts, all examples will be given here both in admittance-type and impedance-type analogies. Table 3.3 shows a comparison of the symbolic representation of elements in each analogy.

Table 3.3

Conversion from admittance-type analogy to impedance-type analogy, or vice versa
Element	Mechanical analogies		Acoustical analogies
Element	Admittance type	Impedance type	Admittance type	Impedance type
Infinite impedance generator (impedance analogy) and zero admittance generator (admittance analogy)
Zero impedance generator (impedance analogy) and infinite admittance generator (admittance analogy)
Dissipative element – resistance (impedance analogy) and conductance (admittance analogy)
Mass element
Compliant element
Impedance element (impedance analogy) and admittance element (admittance analogy)
Table Continued

Element	Mechanical analogies		Acoustical analogies
Element	Admittance type	Impedance type	Admittance type	Impedance type
Transformation element If there are any drug dosages in this chapter, please verify them and indicate that you have done so by initialing this query. converts from one impedance to another and is useful for coupling between electrical, mechanical or acoustical domains	Mechanical to acoustic (admittance type)		Mechanical to acoustic (impedance type)
Gyration element – converts an admittance circuit to an impedance one or vice versa and is useful for coupling between electrical, mechanical or acoustical domains	Mechanical (admittance) to acoustic (impedance)		Mechanical (impedance) to acoustic (admittance)

Mechanical impedance Z _M and mechanical admittance Y _M

The mechanical impedance is the complex ratio of force to velocity at a given point in a mechanical device. We commonly use the symbol Z _M for mechanical impedance, where the subscript M stands for “mechanical.” The unit is N s/m or mechanical ohm.

In the admittance analogy, the mechanical admittance is the inverse of the mechanical impedance. It may also be referred to as the mechanical mobility, but we shall use the more commonly used term admittance. It is the complex ratio of velocity to force at a given point in a mechanical device. We commonly use the symbol Y _M for mechanical admittance. The unit is m/ N·s or mechanical siemens (S).

Mass M _M

Mass is that physical quantity which when acted on by a force is accelerated in direct proportion to that force. The unit is kg. At first sight, mass appears to be a one-terminal quantity because only one connection is needed to set it in motion. However, the force acting on a mass and the resultant acceleration are reckoned with respect to the earth (inertial frame) so that in reality the second terminal of mass is the earth.

The mechanical symbol used to represent mass is shown in Fig. 3.1. The upper end of the mass moves with a velocity

\tilde{u}

$\tilde{u}$

with respect to the ground. The ⌋-shaped configuration represents the “second” terminal of the mass and has zero velocity. The force can be measured by a suitable device inserted between the point 1 and the next element or generator connecting to it.

Mass M _M obeys Newton's second law that

f (t) = M_{M} \frac{d u (t)}{d t},

$f (t) = M_{M} \frac{d u (t)}{d t},$

(3.1)

where f(t) is the instantaneous force in N, M _M is the mass in kg, and u(t) is the instantaneous velocity in m/s.

In the steady state (see Eqs. (2.36–2.44)), with an angular frequency ω equal to 2π times the frequency of vibration, we have the special case of Newton's second law,

\tilde{f} = j ω M_{M} \tilde{u},

$\tilde{f} = j ω M_{M} \tilde{u},$

(3.2)

where

j = \sqrt{- 1}

$j = \sqrt{- 1}$

as usual.

Figure 3.1 Mechanical symbol for a mass.

Figure 3.2 (a) Admittance-type and (b) impedance-type symbols for a mass.

The admittance-type analogous symbol that we use as a replacement for the mechanical symbol in our circuits is a capacitance type, which is shown in Fig. 3.2(a). The mathematical operation invariant for this symbol is found from Table 3.1. In the steady state, we have

\tilde{a} = \frac{\tilde{b}}{j ω c} or \tilde{u} = \frac{\tilde{f}}{j ω M_{M}} .

$\tilde{a} = \frac{\tilde{b}}{j ω c} or \tilde{u} = \frac{\tilde{f}}{j ω M_{M}} .$

(3.3)

This equation is seen to satisfy the physical law given in Eq. (3.2). Note the similarity in appearance of the mechanical and analogous symbols in Figs. 3.1 and 3.2(a). In electrical circuits, the time integral of the current through a capacitor is charge. The analogous quantity here is the time integral of force, which is momentum.

The impedance-type analogous symbol for a mass is an inductance, which is shown in Fig. 3.2(b). The invariant operation for steady state is

\tilde{a} = j ω c \tilde{b}

$\tilde{a} = j ω c \tilde{b}$

\tilde{f} = j ω M_{M} \tilde{u}

$\tilde{f} = j ω M_{M} \tilde{u}$

. It also satisfies Eq. (3.2). Note, however, that in this analogy, one side of the mass element is not necessarily grounded; this often leads to confusion. In electrical circuits, the time integral of the voltage across an inductance is flux turns. The analogous quantity here is momentum.

Mechanical compliance C _M

A physical structure is said to be a mechanical compliance C _M if, when it is acted on by a force, it is displaced in direct proportion to the force. The unit is m/N. Compliant elements usually have two apparent terminals.

The mechanical symbol used to represent a mechanical compliance is a spring. It is shown in Fig. 3.3. The upper end of the element moves with a velocity

{\tilde{u}}_{1}

${\tilde{u}}_{1}$

and the lower end with a velocity

{\tilde{u}}_{2}

${\tilde{u}}_{2}$

. The force required to produce the difference between the velocities

{\tilde{u}}_{1}

${\tilde{u}}_{1}$

and

{\tilde{u}}_{2}

${\tilde{u}}_{2}$

may be measured by breaking into the machine at either point 1 or point 2. Just as the same current would be measured at either end of an element in an electrical circuit, the same force will be found here at either end of the compliant element.

Figure 3.3 Mechanical symbol for a mechanical compliance.

Mechanical compliance C _M obeys the following physical law:

a = \frac{1}{c} \int b d t or f (t) = \frac{1}{C_{M}} \int u (t) d t,

$a = \frac{1}{c} \int b d t or f (t) = \frac{1}{C_{M}} \int u (t) d t,$

(3.4)

where C _M is the mechanical compliance in m/N, and u(t) is the instantaneous velocity in m/s equal to

{\tilde{u}}_{1} - {\tilde{u}}_{2}

${\tilde{u}}_{1} - {\tilde{u}}_{2}$

, the difference in velocity of the two ends.

In the steady state, with an angular frequency ω, equal to 2π times the frequency of vibration, we have

\tilde{f} = \frac{\tilde{u}}{j ω C_{M}},

$\tilde{f} = \frac{\tilde{u}}{j ω C_{M}},$

(3.5)

where

\tilde{f}

$\tilde{f}$

and

\tilde{u}

$\tilde{u}$

are taken to be complex quantities.

The admittance-type analogous symbol used as a replacement for the mechanical symbol in our circuits is an inductance, which is shown in Fig. 3.4(a). The invariant mathematical operation that this symbol represents is given in Table 3.1. In the steady state, we have

\tilde{u} = j ω C_{M} \tilde{f} .

$\tilde{u} = j ω C_{M} \tilde{f} .$

(3.6)

Figure 3.4 (a) Admittance-type and (b) impedance-type symbols for a mechanical compliance.

In electrical circuits, the time integral of the voltage across an inductance is flux turns. The analogous quantity here is the time integral of velocity, which is displacement.

This equation satisfies the physical law given in Eq. (3.5). Note the similarity in appearance of the mechanical and analogous symbols in Fig. 3.3 and 3.4(a).

The impedance-type analogous symbol for a mechanical compliance is a capacitance. It is shown in Fig. 3.4(b). The invariant operation for steady state is

\tilde{a} = \tilde{b} / j ω c

$\tilde{a} = \tilde{b} / j ω c$

\tilde{f} = \tilde{u} / j ω C_{M}

$\tilde{f} = \tilde{u} / j ω C_{M}$

. It also satisfies Eq. (3.5). In electrical circuits, the time integral of the current through a capacitor is the charge. The analogous quantity here is the displacement.

Mechanical resistance R _M and mechanical conductance G _M

A physical structure is said to be a mechanical resistance R _M if, when it is acted on by a force, it moves with a velocity directly proportional to the force. The unit is N·s/m or rayls m².

We also define here a quantity G _M, the mechanical conductance, which is the reciprocal of R _M. The unit of conductance is m/N·s or 1/rayls m².

The above representation for mechanical resistance is usually limited to viscous resistance. Frictional resistance is excluded because, for it, the ratio of force to velocity is not a constant. Both terminals of resistive elements can usually be located by visual inspection.

The mechanical element used to represent viscous resistance is the fluid dashpot shown schematically in Fig. 3.5. The upper end of the element moves with a velocity

{\tilde{u}}_{1}

${\tilde{u}}_{1}$

and the lower with a velocity

{\tilde{u}}_{2}

${\tilde{u}}_{2}$

. The force required to produce the difference between the two velocities

{\tilde{u}}_{1}

${\tilde{u}}_{1}$

and

{\tilde{u}}_{2}

${\tilde{u}}_{2}$

may be measured by breaking into the machine at either point 1 or point 2.

Mechanical resistance R _M obeys the following physical law:

\tilde{f} = R_{M} \tilde{u} = \frac{1}{G_{M}} \tilde{u},

$\tilde{f} = R_{M} \tilde{u} = \frac{1}{G_{M}} \tilde{u},$

(3.7)

where

\tilde{f}

$\tilde{f}$

is the force in N,

\tilde{u}

$\tilde{u}$

is the difference between the velocities

{\tilde{u}}_{1}

${\tilde{u}}_{1}$

and

{\tilde{u}}_{2}

${\tilde{u}}_{2}$

of the two ends, R _M is the mechanical resistance in N·s/m, and G _M is the mechanical conductance in m/N s.

Figure 3.5 Mechanical symbol for mechanical (viscous) resistance.

Figure 3.6 (a) Admittance-type and (b) impedance-type symbols for a mechanical resistance.

The admittance-type analogous symbol used to replace the mechanical symbol in our circuits is a resistance. It is shown in Fig. 3.6(a). The invariant mathematical operation that this symbol represents is given in Table 3.1. In either the steady or transient state, we have

\tilde{u} = G_{M} \tilde{f} = \frac{1}{R_{M}} \tilde{f} .

$\tilde{u} = G_{M} \tilde{f} = \frac{1}{R_{M}} \tilde{f} .$

(3.8)

In the steady state,

\tilde{u}

$\tilde{u}$

and

\tilde{f}

$\tilde{f}$

are taken to be complex quantities. This equation satisfies the physical law given in Eq. (3.7).

The impedance-type analogous symbol for a mechanical resistance is shown in Fig. 3.6(b). It also satisfies Eq. (3.7).

Mechanical generators

The mechanical generators considered will be one of the two types: constant velocity or constant force. A constant velocity generator is represented as a very strong motor attached to a shuttle mechanism in the manner shown in Fig. 3.7. The opposite ends of the generator have velocities

{\tilde{u}}_{1}

${\tilde{u}}_{1}$

and

{\tilde{u}}_{2}

${\tilde{u}}_{2}$

. One of these velocities, either

{\tilde{u}}_{1}

${\tilde{u}}_{1}$

{\tilde{u}}_{2}

${\tilde{u}}_{2}$

, is determined by factors external to the generator. The difference between the velocities

{\tilde{u}}_{1}

${\tilde{u}}_{1}$

and

{\tilde{u}}_{2}

${\tilde{u}}_{2}$

, however, is a velocity

\tilde{u}

$\tilde{u}$

that is independent of the external load connected to the generator.

Figure 3.7 Mechanical symbol for a constant velocity generator.

Figure 3.8 (a) Admittance-type and (b) impedance-type symbols for a constant velocity generator.

The symbols that we used in the two analogies to replace the mechanical symbol for a constant velocity generator are shown in Fig. 3.8. The invariant mathematical operations that these symbols represent are also given in Table 3.1. The tips of the arrows point to the “positive” terminals of the generators. The wave inside the circle in Fig. 3.8(a) indicates that the internal admittance of the generator is zero. The arrow inside the circle of Fig. 3.8(b) indicates that the internal impedance of the generator is infinite.

A constant force generator is represented here by an electromagnetic transducer (e.g., a moving-coil loudspeaker) in the primary of which an electric current of constant amplitude is maintained. Such a generator produces a force equal to the product of the current

\tilde{i}

$\tilde{i}$

, the flux density B, and the effective length of the wire l cutting the flux

(\tilde{f} = B l \tilde{i})

$(\tilde{f} = B l \tilde{i})$

. This device is shown schematically in Fig. 3.9. The opposite ends of the generator have velocities

{\tilde{u}}_{1}

${\tilde{u}}_{1}$

and

{\tilde{u}}_{2}

${\tilde{u}}_{2}$

that are determined by factors external to the generator. The force that the generator produces and that may be measured by breaking into the device at either point 1 or point 2 is a constant force, independent of what is connected to the generator.

The symbols used in the two analogies to replace the mechanical symbol for a constant force generator are given in Fig. 3.10. The invariant mathematical operations that these symbols represent are also given in Table 3.1. The arrows point in the direction of positive flow. Here the arrow inside the circle indicates infinite admittance and the wave inside the circle zero impedance.

Figure 3.9 Mechanical symbol for a constant force generator.

Figure 3.10 (a) Admittance-type and (b) impedance-type symbols for a constant force generator.

Levers

Simple lever

It is apparent that the lever is a device closely analogous to a transformer. The lever in its simplest form consists of a weightless bar resting on an immovable fulcrum, so arranged that a downward force on one end causes an upward force on the other end (see Fig. 3.11). From elementary physics, we may write the equation of balance of moments around the fulcrum

{\tilde{f}}_{1} l_{1} = {\tilde{f}}_{2} l_{2}

${\tilde{f}}_{1} l_{1} = {\tilde{f}}_{2} l_{2}$

or, if not balanced, assuming small displacements,

{\tilde{u}}_{1} l_{2} = {\tilde{u}}_{2} l_{1} .

${\tilde{u}}_{1} l_{2} = {\tilde{u}}_{2} l_{1} .$

(3.9)

Also,

\begin{matrix} Y_{M 1} = \frac{{\tilde{u}}_{1}}{{\tilde{f}}_{1}} = {(\frac{l_{1}}{l_{2}})}^{2} Y_{M 2}, \\ Z_{M 1} = \frac{\tilde{f}}{{\tilde{u}}_{1}} = {(\frac{l_{2}}{l_{1}})}^{2} Z_{M 2} . \end{matrix}

$\begin{matrix} Y_{M 1} = \frac{{\tilde{u}}_{1}}{{\tilde{f}}_{1}} = {(\frac{l_{1}}{l_{2}})}^{2} Y_{M 2}, \\ Z_{M 1} = \frac{\tilde{f}}{{\tilde{u}}_{1}} = {(\frac{l_{2}}{l_{1}})}^{2} Z_{M 2} . \end{matrix}$

(3.10)

The above equations may be represented by the ideal transformers of Fig. 3.12, having a transformation ratio of (l ₁/l ₂):1 for the admittance type and (l ₂/l ₁):1 for the impedance type.

Figure 3.12 (a) Admittance-type and (b) impedance-type symbols for a simple lever.

Floating lever

As an example of a simple floating lever, consider a weightless bar resting on a fulcrum that yields under force. The bar is so arranged that a downward force on one end tends to produce an upward force on the other end. An example is shown in Fig. 3.13.

To solve this type of problem, we first write the equations of moments. Summing the moments about the center support gives

l_{1} {\tilde{f}}_{1} = l_{2} {\tilde{f}}_{2}

$l_{1} {\tilde{f}}_{1} = l_{2} {\tilde{f}}_{2}$

and summing the moments about the end support gives

(l_{1} + l_{2}) {\tilde{f}}_{1} = l_{2} {\tilde{f}}_{3} .

$(l_{1} + l_{2}) {\tilde{f}}_{1} = l_{2} {\tilde{f}}_{3} .$

(3.11)

When the forces are not balanced and if we assume infinitesimal displacements, the velocities are related to the forces through the admittances so that

\begin{matrix} {\tilde{u}}_{3} = Y_{M 3} {\tilde{f}}_{3} = Y_{M 3} \frac{l_{1} + l_{2}}{l_{2}} {\tilde{f}}_{1}, \\ {\tilde{u}}_{2} = Y_{M 2} {\tilde{f}}_{2} = Y_{M 2} \frac{l_{1}}{l_{2}} {\tilde{f}}_{1} . \end{matrix}

$\begin{matrix} {\tilde{u}}_{3} = Y_{M 3} {\tilde{f}}_{3} = Y_{M 3} \frac{l_{1} + l_{2}}{l_{2}} {\tilde{f}}_{1}, \\ {\tilde{u}}_{2} = Y_{M 2} {\tilde{f}}_{2} = Y_{M 2} \frac{l_{1}}{l_{2}} {\tilde{f}}_{1} . \end{matrix}$

(3.12)

Also, by superposition, it is seen from simple geometry that

Figure 3.14 Admittance-type symbol for a floating lever.

{\tilde{u}}_{1}^{'} = {\tilde{u}}_{3} \frac{l_{1} + l_{2}}{l_{2}} for {\tilde{u}}_{2} = 0,

${\tilde{u}}_{1}^{'} = {\tilde{u}}_{3} \frac{l_{1} + l_{2}}{l_{2}} for {\tilde{u}}_{2} = 0,$

{\tilde{u}}_{1}^{″} = {\tilde{u}}_{2} \frac{l_{1}}{l_{2}} for {\tilde{u}}_{3} = 0,

${\tilde{u}}_{1}^{″} = {\tilde{u}}_{2} \frac{l_{1}}{l_{2}} for {\tilde{u}}_{3} = 0,$

so that

{\tilde{u}}_{1} = {\tilde{u}}_{1}^{'} + {\tilde{u}}_{1}^{″} = \frac{l_{1} + l_{2}}{l_{2}} {\tilde{u}}_{3} + \frac{l_{1}}{l_{2}} {\tilde{u}}_{2}

${\tilde{u}}_{1} = {\tilde{u}}_{1}^{'} + {\tilde{u}}_{1}^{″} = \frac{l_{1} + l_{2}}{l_{2}} {\tilde{u}}_{3} + \frac{l_{1}}{l_{2}} {\tilde{u}}_{2}$

(3.13)

and, finally,

\frac{{\tilde{u}}_{1}}{{\tilde{f}}_{1}} = Y_{M 1} = Y_{M 3} {(\frac{l_{1} + l_{2}}{l_{2}})}^{2} + Y_{M 2} {(\frac{l_{1}}{l_{2}})}^{2} .

$\frac{{\tilde{u}}_{1}}{{\tilde{f}}_{1}} = Y_{M 1} = Y_{M 3} {(\frac{l_{1} + l_{2}}{l_{2}})}^{2} + Y_{M 2} {(\frac{l_{1}}{l_{2}})}^{2} .$

(3.14)

This equation may be represented by the analogous circuit of Fig. 3.14. The lever loads the generator with two admittances connected in series, each of which behaves as a simple lever when the other is equal to zero. It will be seen that this is a way of obtaining the equivalent of two series masses without a common zero-velocity (ground) point. This will be illustrated in Example 3.3.

Figure 3.15 Six-element mechanical device.

Example 3.1. The mechanical device of Fig. 3.15 consists of a piston of mass M _M1 sliding on an oil surface inside a cylinder of mass M _M2. This cylinder in turn slides in an oiled groove cut in a rigid body. The sliding (viscous) resistances are R _M1 and R _M2, respectively. The cylinder is held by a spring of compliance C _M. The mechanical generator maintains a constant sinusoidal velocity of angular frequency ω, whose magnitude is

\tilde{u}

$\tilde{u}$

m/s. Solve for the force

\tilde{f}

$\tilde{f}$

produced by the generator.

Solution. Although the force will be determined ultimately from an analysis of the admittance-type analogous circuit for this mechanical device, it is frequently useful to draw a mechanical circuit diagram. This interim step to the desired circuit will be especially helpful to the student who is inexperienced in the use of analogies. Its use virtually eliminates errors from the final circuit.

To draw the mechanical circuit, note first the junction points of two or more elements. This locates all element terminals which move with the same velocity. There are in this example two velocities,

\tilde{u}

$\tilde{u}$

and

{\tilde{u}}_{2}

${\tilde{u}}_{2}$

, in addition to “ground” or zero velocity. These two velocities are represented in the mechanical circuit diagram by the velocities of two imaginary rigid bars, 1 and 2 of Fig. 3.16, which oscillate in a vertical direction. The circuit drawing is made by attaching all element terminals with velocity

\tilde{u}

$\tilde{u}$

to the first bar and all terminals with velocity

{\tilde{u}}_{2}

${\tilde{u}}_{2}$

to the second bar. All terminals with zero velocity are drawn to a ground bar. Note that a mass always has one terminal on ground [8]. Three elements of Fig. 3.15 have one terminal with the velocity

\tilde{u}

$\tilde{u}$

: the generator, the mass M _M1, and the viscous resistance R _M1. These are attached to bar 1. Four elements have one terminal with the velocity

{\tilde{u}}_{2}

${\tilde{u}}_{2}$

: the viscous resistances R _M1 and R _M2, the mass M _M2, and the compliance C _M. These are attached to bar 2. Five elements have one terminal with zero velocity: the generator, both masses, the viscous resistance R _M2, and the compliance C _M.

We are now in a position to transform the mechanical circuit into an admittance-type analogous circuit. This is accomplished simply by replacing the mechanical elements with the analogous admittance-type elements. The circuit becomes such as that shown in Fig. 3.17. Remember that, in the admittance-type analogy, force “flows” through the elements and velocity is the drop across them. The resistors must have Gs written alongside them. As defined above, G _M = 1/R _M, and the unit is m/N·s or mechanical siemens.

Figure 3.16 Mechanical circuit for the device of Fig. 3.15.

Figure 3.17 Admittance-type analogous circuit for the device of Fig. 3.15.

The equations for this circuit are found in the usual manner, using the rules of Table 3.1. Let us determine

Y_{M} = \tilde{u} / \tilde{f}

$Y_{M} = \tilde{u} / \tilde{f}$

, the mechanical admittance presented to the generator. The mechanical admittance of the three elements in parallel on the right-hand side of the schematic diagram is

\begin{matrix} \frac{{\tilde{u}}_{2}}{{\tilde{f}}_{2}} = \frac{1}{\frac{1}{1 / j ω M_{M 2}} + \frac{1}{G_{M 2}} + \frac{1}{j ω C_{M}}} \\ = \frac{1}{j ω M_{M 2} + R_{M 2} + \frac{1}{j ω C_{M}}} . \end{matrix}

$\begin{matrix} \frac{{\tilde{u}}_{2}}{{\tilde{f}}_{2}} = \frac{1}{\frac{1}{1 / j ω M_{M 2}} + \frac{1}{G_{M 2}} + \frac{1}{j ω C_{M}}} \\ = \frac{1}{j ω M_{M 2} + R_{M 2} + \frac{1}{j ω C_{M}}} . \end{matrix}$

Including the element G _M1, the mechanical admittance for that part of the circuit through which

{\tilde{f}}_{2}

${\tilde{f}}_{2}$

flows is, then,

\frac{\tilde{u}}{{\tilde{f}}_{2}} = G_{M 1} + \frac{1}{j ω M_{M 2} + R_{M 2} + \frac{1}{j ω C_{M}}} .

$\frac{\tilde{u}}{{\tilde{f}}_{2}} = G_{M 1} + \frac{1}{j ω M_{M 2} + R_{M 2} + \frac{1}{j ω C_{M}}} .$

Note that the input mechanical admittance Y _M is given by

Y_{M} = \frac{\tilde{u}}{\tilde{f}} = \frac{\tilde{u}}{{\tilde{f}}_{1} + {\tilde{f}}_{2}} .

$Y_{M} = \frac{\tilde{u}}{\tilde{f}} = \frac{\tilde{u}}{{\tilde{f}}_{1} + {\tilde{f}}_{2}} .$

and

{\tilde{f}}_{1} = \frac{\tilde{u}}{1 / j ω M_{M 1}} = j ω M_{M 1} \tilde{u} .

${\tilde{f}}_{1} = \frac{\tilde{u}}{1 / j ω M_{M 1}} = j ω M_{M 1} \tilde{u} .$

Substituting

{\tilde{f}}_{1}

${\tilde{f}}_{1}$

and

{\tilde{f}}_{2}

${\tilde{f}}_{2}$

into the second equation preceding gives us the input admittance:

Y_{M} = \frac{\tilde{u}}{\tilde{f}} = \frac{1}{j ω M_{M 1} + \frac{1}{G_{M 1} + \frac{1}{j ω M_{M 2} + R_{M 2} + \frac{1}{j ω C_{M}}}}} .

$Y_{M} = \frac{\tilde{u}}{\tilde{f}} = \frac{1}{j ω M_{M 1} + \frac{1}{G_{M 1} + \frac{1}{j ω M_{M 2} + R_{M 2} + \frac{1}{j ω C_{M}}}}} .$

(3.15a)

Figure 3.18 Three-element mechanical device.

The mechanical impedance is the reciprocal of Eq. (3.15a):

Z_{M} = \frac{\tilde{f}}{\tilde{u}} = j ω M_{M 1} + \frac{1}{G_{M 1} + \frac{1}{j ω M_{M 2} + R_{M 2} + \frac{1}{j ω C_{M}}}} .

$Z_{M} = \frac{\tilde{f}}{\tilde{u}} = j ω M_{M 1} + \frac{1}{G_{M 1} + \frac{1}{j ω M_{M 2} + R_{M 2} + \frac{1}{j ω C_{M}}}} .$

(3.15b)

The result is

\tilde{f} = Z_{M} \tilde{u} N .

$\tilde{f} = Z_{M} \tilde{u} N .$

(3.16)

Example 3.2. As a further example of a mechanical circuit, let us consider the two masses of 2 and 4 kg shown in Fig. 3.18. They are assumed to rest on a frictionless plane surface and to be connected together through a generator of constant velocity that is also free to slide on the frictionless plane surface.

Let its velocity be

u_{0} (t) = 2 \cos 1000 t cm / s

$u_{0} (t) = 2 \cos 1000 t cm / s$

so that

{\tilde{u}}_{0} = 2 e^{j 1000 t} cm / s

${\tilde{u}}_{0} = 2 e^{j 1000 t} cm / s$

| {\tilde{u}}_{0} | = 2 cm / s at ω = 1000 Hz .

$| {\tilde{u}}_{0} | = 2 cm / s at ω = 1000 Hz .$

Draw the admittance-type analogous circuit and determine the force

\tilde{f}

$\tilde{f}$

produced by the generator. Also, determine the admittance presented to the generator.

Solution. The masses do not have the same velocity with respect to ground. The difference between the velocities of the two masses is

{\tilde{u}}_{0}

${\tilde{u}}_{0}$

. The element representing a mass is that shown in Fig. 3.2(a) with one end grounded and the other moving at the velocity of the mass.

The admittance-type circuit for this example is shown in Fig. 3.19. The velocity

{\tilde{u}}_{0}

${\tilde{u}}_{0}$

equals

{\tilde{u}}_{1} + {\tilde{u}}_{2}

${\tilde{u}}_{1} + {\tilde{u}}_{2}$

, where

{\tilde{u}}_{1}

${\tilde{u}}_{1}$

is the velocity with respect to ground of M ₁ and

{\tilde{u}}_{2}

${\tilde{u}}_{2}$

is that for M _M3. The force

\tilde{f}

$\tilde{f}$

Figure 3.19 Admittance-type analogous circuit for the device of Fig. 3.18.

\begin{matrix} \tilde{f} = \frac{1}{\frac{1}{j ω M_{M 1}} + \frac{1}{j ω M_{M 2}}} | {\bar{u}}_{0} | e^{j 1000 t} \\ = \frac{j ω M_{M 1} M_{M 2}}{M_{M 1} + M_{M 2}} | {\tilde{u}}_{0} | e^{j 1000 t} \\ = \frac{j 1000 \times 2 \times 4 \times 0.02}{2 + 4} e^{j 1000 t} = j 26.7 e^{j 1000 t} N . \end{matrix}

$\begin{matrix} \tilde{f} = \frac{1}{\frac{1}{j ω M_{M 1}} + \frac{1}{j ω M_{M 2}}} | {\bar{u}}_{0} | e^{j 1000 t} \\ = \frac{j ω M_{M 1} M_{M 2}}{M_{M 1} + M_{M 2}} | {\tilde{u}}_{0} | e^{j 1000 t} \\ = \frac{j 1000 \times 2 \times 4 \times 0.02}{2 + 4} e^{j 1000 t} = j 26.7 e^{j 1000 t} N . \end{matrix}$

(3.17)

The j indicates that the time phase of the force is 90 degrees leading with respect to that of the velocity of the generator. Hence, the rms force is

f_{rms} = \frac{| \tilde{f} |}{\sqrt{2}} ∠ 90^{\circ} = 18.9 N ∠ 90^{\circ}

$f_{rms} = \frac{| \tilde{f} |}{\sqrt{2}} ∠ 90^{\circ} = 18.9 N ∠ 90^{\circ}$

(3.18)

Obviously, when one mass is large compared with the other, the force is that necessary to move the smaller one alone. This example reveals the only type of case in which masses can be in series without the introduction of floating levers. At most, only two masses can be in series because a common ground is necessary.

The admittance presented to the generator is

\begin{matrix} Y_{M} = \frac{{\tilde{u}}_{0}}{\tilde{f}} = \frac{M_{M 1} + M_{M 2}}{j ω M_{M 1} M_{M 2}} \\ = \frac{6}{j 1000 \times 8} = - j 7.5 \times 10^{- 4} m / N \cdot s \end{matrix}

$\begin{matrix} Y_{M} = \frac{{\tilde{u}}_{0}}{\tilde{f}} = \frac{M_{M 1} + M_{M 2}}{j ω M_{M 1} M_{M 2}} \\ = \frac{6}{j 1000 \times 8} = - j 7.5 \times 10^{- 4} m / N \cdot s \end{matrix}$

(3.19)

Example 3.3. An example of a mechanical device embodying a floating lever is shown in Fig. 3.20. The masses attached at points 2 and 3 may be assumed to be resting on very compliant springs. The driving force

{\tilde{f}}_{1}

${\tilde{f}}_{1}$

will be assumed to have a frequency well above the resonance frequencies of the masses and their spring supports so that

Figure 3.20 (a) Mechanical device embodying a floating lever. (b) Mechanical diagram of (a). The compliances of the springs are very large so that all of f ₂ and f ₃ go to move M _M2 and M _M3.

\begin{matrix} Y_{M 2} \approx \frac{1}{j ω M_{M 2}} \\ Y_{M 3} \approx \frac{1}{j ω M_{M 3}} \end{matrix}

$\begin{matrix} Y_{M 2} \approx \frac{1}{j ω M_{M 2}} \\ Y_{M 3} \approx \frac{1}{j ω M_{M 3}} \end{matrix}$

Also, assume that a mass is attached to the weightless lever bar at point 1, with an admittance

Y_{M 1} = \frac{1}{j ω M_{M 1}} .

$Y_{M 1} = \frac{1}{j ω M_{M 1}} .$

Solve for the total admittance presented to the constant force generator

{\tilde{f}}_{1}

${\tilde{f}}_{1}$

Solution. By inspection, the admittance-type analogous circuit is drawn as shown in Fig. 3.21(a,b). Solving for

Y_{M} = {\tilde{u}}_{1} / {\tilde{f}}_{1}

$Y_{M} = {\tilde{u}}_{1} / {\tilde{f}}_{1}$

, we get

Y_{M} = \frac{1}{j ω [\frac{M_{M 2} M_{M 3} l_{2}^{2}}{M_{M 3} l_{1}^{2} + M_{M 2} {(l_{1} + l_{2})}^{2}} + M_{M 1}]}

$Y_{M} = \frac{1}{j ω [\frac{M_{M 2} M_{M 3} l_{2}^{2}}{M_{M 3} l_{1}^{2} + M_{M 2} {(l_{1} + l_{2})}^{2}} + M_{M 1}]}$

(3.20)

Note that if l ₂ → 0, the admittance is simply that of the mass M _M1. Also, if l ₁ → 0, the admittance is that of M _M1 and M _M3, that is,

Figure 3.21 (a) Admittance-type analogous circuit for the device of Fig. 3.20. (b) Same as (a) but with transformers removed.

Y_{M} = \frac{1}{j ω (M_{M 3} + M_{M 1})}

$Y_{M} = \frac{1}{j ω (M_{M 3} + M_{M 1})}$

(3.21)

In an admittance-type circuit (with transformers eliminated), it is possible with one or more floating levers to have one or more M _Ms with no ground terminal (s).

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Table of Contents for Acoustics

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3.3. Mechanical elements

Mechanical impedance Z M and mechanical admittance Y M

Mass M M

Mechanical compliance C M

Mechanical resistance R M and mechanical conductance G M

Mechanical generators

Levers

Simple lever

Floating lever

Table of Contents for
Acoustics

Mechanical impedance Z _M and mechanical admittance Y _M

Mass M _M

Mechanical compliance C _M

Mechanical resistance R _M and mechanical conductance G _M