FORCED VIBRATION UNDER HARMONIC LOADING
1 Response of Undamped SDOF System to Harmonic Motion
In the forced vibration system, due to periodic force applied to the system, the vibrations continue as long
as the periodic force is present. Periodic force which is harmonic in nature will be considered.
𝑓(𝑡) = 𝑓𝑜 𝑠𝑖𝑛𝜔𝑓 𝑡
Where 𝑓𝑜 is the peak amplitude and 𝜔𝑓 is the frequency of the force in radians per second.
Thus
𝑚
𝑑2 𝑥
+ 𝑘𝑥 = 𝑓𝑜 𝑠𝑖𝑛𝜔𝑓 𝑡
𝑑𝑡 2
So the differential equation of the periodic motions has a solution.
𝑥𝑐 = 𝑥(𝑡) = 𝑒 𝛼𝑡 (𝐴1 𝑐𝑜𝑠𝛽𝑡 + 𝐴2 𝑠𝑖𝑛𝛽𝑡) = 𝐴1 𝑐𝑜𝑠𝜔𝑡 + 𝐴2 𝑠𝑖𝑛𝜔𝑡
Particular solution
𝑥𝑝 = 𝑋𝑠𝑖𝑛𝜔𝑓 𝑡
𝑑𝑥𝑝
= 𝑋𝜔𝑓 𝑐𝑜𝑠𝜔𝑓 𝑡
𝑑𝑡
𝑑2 𝑥𝑝
= −𝑋𝜔𝑓 2 𝑠𝑖𝑛𝜔𝑓 𝑡
𝑑𝑡 2
Where 𝑋 is the peak value of a particular solution
Substituting
−𝑚𝑋𝜔𝑓 2 𝑠𝑖𝑛𝜔𝑓 𝑡 + 𝑘𝑋𝑠𝑖𝑛𝜔𝑓 𝑡 = 𝑓𝑜 𝑠𝑖𝑛𝜔𝑓 𝑡
−𝑚𝑋𝜔𝑓 2 + 𝑘𝑋 = 𝑓𝑜
𝑋=
𝑓𝑜
𝑘 − 𝑚𝜔𝑓 2
𝑓𝑜
𝑓𝑜
𝑘
𝑘
𝑋=
=
𝑚𝜔𝑓 2
𝜔𝑓 2
1−
1− 2
𝑘
𝜔
Where 𝑟 =
𝜔𝑓 2
𝜔2
is the frequency ratio of the applied force frequency to natural frequency of vibration of the
system.
Therefore
𝑓𝑜
𝑋= 𝑘 2
1−𝑟
1
Therefore the final solution will be
𝑓𝑜
𝑥(𝑡) = 𝐴1 𝑐𝑜𝑠𝜔𝑡 + 𝐴2 𝑠𝑖𝑛𝜔𝑡 + 𝑘 2 𝑠𝑖𝑛𝜔𝑓 𝑡
1−𝑟
Initial conditions are applied to obtain constant𝐴1 𝑎𝑛𝑑𝐴2 . (Recall: differential equations)
In the case of initial condition
Considering initial conditions 𝑡 = 0 𝑥(0) = 0
𝑣=0
𝑓𝑜
𝑘
𝐴2 = −
1 − 𝑟2
𝑟
𝐴1 = 0,
Thus
𝑓𝑜
𝑓𝑜
𝑘 𝑠𝑖𝑛𝜔𝑡 + 𝑘 𝑠𝑖𝑛𝜔 𝑡
𝑥(𝑡) = −
𝑓
1 − 𝑟2
1 − 𝑟2
𝑟
𝑓𝑜
𝑥(𝑡) = − 𝑘 2 [𝑠𝑖𝑛𝜔𝑓 𝑡 − 𝑟𝑠𝑖𝑛𝜔𝑡]
1−𝑟
As damping forces will always be present, the free frequency term will vanish eventually. Therefore the
term (𝑟𝑠𝑖𝑛𝜔𝑡) represents transient response of vibration. Transient response of vibration occurs to a system
while it is approaching steady state.
The forcing term is referred to as steady-state response is given by:
𝑓𝑜
𝑥(𝑡) = 𝑘 2 𝑠𝑖𝑛𝜔𝑓 𝑡, 𝑟 ≠ 1
1−𝑟
When r =1, 𝜔𝑓 = 𝜔 t the amplitude of motions becomes infinitely large. This is resonance. The amplitude
will increase gradually to infinite. The materials used in practice have strength limitation, thus structural
failures occur long before extremely large amplitudes can be attained.
Let
𝑓𝑜
= 𝑋0
𝑘
𝑋0 is the static deflection
𝑋=
𝑋0
1 − 𝑟2
𝑋
1
=
= 𝐹(𝜔𝑓 )
𝑋0 1 − 𝑟 2
𝑋
𝐹(𝜔𝑓 ) = 𝑋 is called frequency response function. It gives the magnitude and he sign of steady state motion
0
as a function of frequency ratio “r”
2
The magnitude𝐷𝑠 , is called the steady-state or dynamic magnification factor, or gain as shown in the figure
below
𝐷𝑠 = |𝐹(𝜔𝑓 )|
If r<1, the response is in phase with the excitation, since (1-r2) is positive. If r is more than 1.0, the response
is 180o out of phase with the excitation, that is
𝑓𝑜
𝑥(𝑡) = 𝑘 2 (−𝑠𝑖𝑛𝜔𝑓 𝑡)
1−𝑟
𝑥(𝑡) =
𝑋0
(−𝑠𝑖𝑛𝜔𝑓 𝑡)
1 − 𝑟2
When r is close to one but not equal to 1, the forced motion and natural motion alternatively reinforce each
other and cancel each other giving the appearance of a beat phenomenon.
2 Response of Viscous-Damped SDOF System to Harmonic Motion
Equation of motion
𝑚
𝑑2 𝑥
𝑑𝑥
+𝑐
+ 𝑘𝑥 = 𝑓𝑜 𝑠𝑖𝑛𝜔𝑓 𝑡
2
𝑑𝑡
𝑑𝑡
Solution
𝑥(𝑡) = 𝑥𝑐 (𝑡) + 𝑥𝑝 (𝑡)
3
Where 𝑥𝑐 (𝑡) is the complementary solution and 𝑥𝑝 (𝑡) is the particular solution. For underdamped case
where 𝐶 < 𝐶𝑐𝑟
𝑥𝑐 (𝑡) = 𝑒 −𝜉𝜔𝑡 (𝐴1 𝑐𝑜𝑠𝜔𝐷 𝑡 + 𝐴2 𝑠𝑖𝑛𝜔𝐷 𝑡)
Particular solution is given by
𝑓(𝑡) = 𝑠𝑖𝑛𝜔𝑓 𝑡
𝑑𝑓(𝑡)
𝑑𝑡
𝑑2 𝑓(𝑡)
2
𝑑𝑡
= 𝜔𝑓 𝑐𝑜𝑠𝜔𝑓 𝑡
= −𝜔𝑓 2 𝑠𝑖𝑛𝜔𝑓 𝑡
Gives two independent function 𝑠𝑖𝑛𝜔𝑓 𝑡 𝑎𝑛𝑑 𝑐𝑜𝑠𝜔𝑓 𝑡
𝑥𝑝 = 𝑘1 𝑠𝑖𝑛𝜔𝑓 𝑡 + 𝑘2 𝑐𝑜𝑠𝜔𝑓 𝑡
𝑑𝑥𝑝
𝑑𝑡
𝑑2 𝑥𝑝
2
𝑑𝑡
Substituting into𝑚
𝑑2 𝑥
𝑑𝑡 2
+𝑐
𝑑𝑥
𝑑𝑡
= 𝑘1 𝜔𝑓 𝑐𝑜𝑠𝜔𝑓 𝑡 − 𝑘2 𝜔𝑓 𝑠𝑖𝑛𝜔𝑓 𝑡
= −𝑘1 𝜔𝑓 2 𝑠𝑖𝑛𝜔𝑓 𝑡 − 𝑘2 𝜔𝑓 2 𝑐𝑜𝑠𝜔𝑓 𝑡
+ 𝑘𝑥 = 𝑓𝑜 𝑠𝑖𝑛𝜔𝑓 𝑡
𝑚(−𝑘1 𝜔𝑓 2 𝑠𝑖𝑛𝜔𝑓 𝑡 − 𝑘2 𝜔𝑓 2 𝑐𝑜𝑠𝜔𝑓 𝑡) + 𝑐(𝑘1 𝜔𝑓 𝑐𝑜𝑠𝜔𝑓 𝑡 − 𝑘2 𝜔𝑓 𝑠𝑖𝑛𝜔𝑓 𝑡) + 𝑘(𝑘1 𝑠𝑖𝑛𝜔𝑓 𝑡 + 𝑘2 𝑐𝑜𝑠𝜔𝑓 𝑡)
= 𝑓𝑜 𝑠𝑖𝑛𝜔𝑓 𝑡
−𝑚𝑘1 𝜔𝑓 2 𝑠𝑖𝑛𝜔𝑓 𝑡 − 𝑚𝑘2 𝜔𝑓 2 𝑐𝑜𝑠𝜔𝑓 𝑡 + 𝑐𝑘1 𝜔𝑓 𝑐𝑜𝑠𝜔𝑓 𝑡 − 𝑐𝑘2 𝜔𝑓 𝑠𝑖𝑛𝜔𝑓 𝑡 + 𝑘𝑘1 𝑠𝑖𝑛𝜔𝑓 𝑡 + k𝑘2 𝑐𝑜𝑠𝜔𝑓 𝑡
= 𝑓𝑜 𝑠𝑖𝑛𝜔𝑓 𝑡
(−𝑚𝑘1 𝜔𝑓 2 − 𝑐𝑘2 𝜔𝑓 + 𝑘𝑘1 )𝑠𝑖𝑛𝜔𝑓 𝑡 + (−𝑚𝑘2𝜔𝑓 2 + 𝑐𝑘1 𝜔𝑓 + k𝑘2 )𝑐𝑜𝑠𝜔𝑓 𝑡 = 𝑓𝑜 𝑠𝑖𝑛𝜔𝑓 𝑡
The equations yields two algebraic equations
−𝑚𝑘1 𝜔𝑓 2 − 𝑐𝑘2 𝜔𝑓 + 𝑘𝑘1 = 𝑓𝑜
(𝑘 − 𝑚 𝜔𝑓 2 )𝑘1 − 𝑐𝜔𝑓 𝑘2 = 𝑓𝑜
−𝑚𝑘2 𝜔𝑓 2 + 𝑐𝑘1 𝜔𝑓 + k𝑘2 = 0
𝑐𝑘1 𝜔𝑓 + (k − 𝑚𝜔𝑓 2 )𝑘2 = 0
Divide through by k
(1 −
𝑚 2
𝑐
𝑓𝑜
𝜔𝑓 )𝑘1 − 𝜔𝑓 𝑘2 =
𝑘
𝑘
𝑘
4
𝑐
𝑚
𝑘1 𝜔𝑓 + (1 − 𝜔𝑓 2 )𝑘2 = 0
𝑘
𝑘
(1 − 𝑟 2 )𝑘1 − 2𝑟𝜉𝑘2 = 𝑋𝑜
2𝑟𝜉𝑘1 + (1 − 𝑟 2 )𝑘2 = 0
𝑐
𝑐
Where 𝜉 = 𝑐 = 2𝑚𝜔 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑐 = 2𝜉𝑚𝜔
𝑐𝑟
𝑘1 =
(1 − 𝑟 2 )𝑋𝑜
(1 − 𝑟 2 )2 + (2𝑟𝜉)2
𝑘2 = −
(2𝑟𝜉)𝑋𝑜
(1 − 𝑟 2 )2 + (2𝑟𝜉)2
Therefore
𝑥𝑝 =
(1 − 𝑟 2 )𝑋𝑜
(2𝑟𝜉)𝑋𝑜
𝑠𝑖𝑛𝜔𝑓 𝑡 −
𝑐𝑜𝑠𝜔𝑓 𝑡
2
2
2
(1 − 𝑟 ) + (2𝑟𝜉)
(1 − 𝑟 2 )2 + (2𝑟𝜉)2
𝑥𝑝 =
(1 −
𝑋𝑜
[(1 − 𝑟 2 )𝑠𝑖𝑛𝜔𝑓 𝑡 − (2𝑟𝜉)𝑐𝑜𝑠𝜔𝑓 𝑡]
+ (2𝑟𝜉)2
𝑟 2 )2
Which can be written as
𝑥𝑝 =
𝑋𝑜
√(1 − 𝑟 2 )2 + (2𝑟𝜉)2
sin(𝜔𝑓 𝑡−∝)
Where ∝ is the phase angle defined by
∝= 𝑡𝑎𝑛−1 (
2𝑟𝜉
)
1 − 𝑟2
Can be written in compact form as
𝑥𝑝 = 𝑋𝑜 𝛽 sin(𝜔𝑓 𝑡−∝)
Where 𝛽 is the magnification factor in the case of damped system.
𝛽=
𝑥𝑝
1
=
𝑋𝑜 √(1 − 𝑟 2 )2 + (2𝑟𝜉)2
Total response
𝑥(𝑡) = 𝑒 −𝜉𝜔𝑡 (𝐴1 𝑐𝑜𝑠𝜔𝐷 𝑡 + 𝐴2 𝑠𝑖𝑛𝜔𝐷 𝑡) +
𝑋𝑜
√(1 −
𝑟 2 )2
+ (2𝑟𝜉)2
sin(𝜔𝑓 𝑡−∝)
The constants 𝐴1 and 𝐴2 are determined from the initial conditions.
The figure below gives the steady-state magnification factor ‘𝛽’ against frequency ratio ‘r’ and phase angle
∝ against frequency ratio “r” for various amounts of damping ratios 𝝃
5
FORCE
3 Transmission
It is clear that by increasing the spring stiffness k and the damping coefficient c, the amplitude of vibration
decreases. The increase in the stiffness and damping coefficients, however, may have an adverse effect on
the force transmitted to the support. In order to reduce the force transmitted to the support, the stiffness
and damping coefficients must be properly selected.
Given the equation of motion
𝑚
𝑑2 𝑥
𝑑𝑥
+𝑐
+ 𝑘𝑥 = 𝑓𝑜 𝑠𝑖𝑛𝜔𝑓 𝑡
2
𝑑𝑡
𝑑𝑡
With particular solution
𝑥𝑝 = 𝑋𝑜 𝛽 sin(𝜔𝑓 𝑡−∝) (𝑟𝑒𝑐𝑎𝑙𝑙? )
6
𝑑𝑥
The force transmitted to the support through the spring is kx and through the damping element𝑐 𝑑𝑡 . Hence
the total force transmitted to foundation
𝑓𝑡 = 𝑘𝑥 + 𝑐
𝑑𝑥
𝑑𝑡
Therefore
𝑥𝑝 = 𝑋𝑜 𝛽 sin(𝜔𝑓 𝑡−∝)
𝑑𝑥𝑝
𝑑𝑡
= 𝑋𝑜 𝛽𝜔𝑓 cos(𝜔𝑓 𝑡−∝)
Therefore
𝑓𝑡 = 𝑘(𝑋𝑜 𝛽 sin(𝜔𝑓 𝑡−∝)) + 𝑐(𝑋𝑜 𝛽𝜔𝑓 cos(𝜔𝑓 𝑡−∝))
𝑓𝑡 = 𝑋𝑜 𝛽[𝑘 sin(𝜔𝑓 𝑡−∝) + 𝑐 𝜔𝑓 cos(𝜔𝑓 𝑡−∝)]
Given that
𝑐𝜔𝑓
)
𝑘
𝜓 = 𝑡𝑎𝑛−1 (
Then
𝑓𝑡 = 𝑋𝑜 𝛽√𝑘2 + (𝑐 𝜔𝑓 )2 sin(𝜔𝑓 𝑡−∝ +𝜓)
𝑋𝑜 =
𝑓𝑡 =
𝑓𝑜
𝑘
𝛽√𝑘2 + (𝑐 𝜔𝑓 )2 sin(𝜔𝑓 𝑡−∝ +𝜓)
𝑓𝑡 = 𝑓𝑜 𝛽√1 + (
𝑐
𝑓𝑜
𝑘
𝑐 𝜔𝑓
𝑘
2
) sin(𝜔𝑓 𝑡−∝ +𝜓)
𝑐
Where 𝜉 = 𝑐 = 2𝑚𝜔 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑐 = 2𝜉𝑚𝜔
𝑐𝑟
2𝜉𝑚𝜔𝜔𝑓
𝑓𝑡 = 𝑓𝑜 𝛽√1 + (
𝑓𝑡 = 𝑓𝑜 𝛽√1 + (
𝑘
2𝜉𝜔𝑓
𝜔
2
) sin(𝜔𝑓 𝑡−∝ +𝜓)
2
) sin(𝜔𝑓 𝑡−∝ +𝜓)
𝑓𝑡 = 𝑓𝑜 𝛽√1 + (2𝜉𝑟)2 sin(𝜔𝑓 𝑡−∝ +𝜓)
7
Maximum force
𝐹𝑡 = 𝑓𝑜 𝛽√1 + (2𝜉𝑟)2
𝐹𝑡 = 𝑓𝑜
√1 + (2𝜉𝑟)2
√(1 − 𝑟2 )2 + (2𝑟𝜉)2
1 + (2𝜉𝑟)2
√
𝐹𝑡 = 𝑓𝑜
(1 − 𝑟2 )2 + (2𝑟𝜉)2
Transmissibility Tr is defined as the ratio between the amplitude of the force transmitted to the foundation
and the amplitude of the force. Therefore
𝑇𝑟 =
𝐹𝑡
𝑓𝑜
1 + (2𝜉𝑟)2
=√
(1 − 𝑟2 )2 + (2𝑟𝜉)2
The transmissibility of motion from the foundation to the structure and the transmissibility of force from
the structure to the foundation is given by the same expression.
4 Response to Support Motion
There are cases where the foundation or support of a structure is subjected to time varying motion. Ground
motion of earthquakes or other excitations such as explosion or dynamic action of machinery causes motion
of structures.
Harmonic motion
𝑧(𝑡) = 𝑧𝑜 𝑠𝑖𝑛𝜔𝑓 𝑡
Where 𝑧𝑜 us the maximum amplitude and 𝜔𝑓 is the frequency of the support motion.
Given the equation of motion
8
𝑚
𝑑2 𝑥
𝑑𝑥
𝑑𝑧
+𝑐
+ 𝑘𝑥 = 𝑐 + 𝑘𝑧
2
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑧(𝑡) = 𝑧𝑜 𝑠𝑖𝑛𝜔𝑓 𝑡
𝑑𝑧
= 𝑧𝑜 𝜔𝑓 𝑐𝑜𝑠𝜔𝑓 𝑡
𝑑𝑡
Substitution
𝑚
𝑑2 𝑥
𝑑𝑥
+𝑐
+ 𝑘𝑥 = 𝑧𝑜 (𝑘𝑠𝑖𝑛𝜔𝑓 𝑡 + 𝑐𝜔𝑓 𝑐𝑜𝑠𝜔𝑓 𝑡)
2
𝑑𝑡
𝑑𝑡
𝑚
𝑑2 𝑥
𝑑𝑥
+𝑐
+ 𝑘𝑥 = 𝑧𝑜 √𝑘 2 + (𝑐𝜔𝑓 )2 sin(𝜔𝑓 𝑡 +∝𝑧 )
2
𝑑𝑡
𝑑𝑡
𝑚
𝑐𝜔𝑓 2
𝑑2 𝑥
𝑑𝑥
+𝑐
+ 𝑘𝑥 = 𝑧𝑜 𝑘√1 + (
) sin(𝜔𝑓 𝑡 +∝𝑧 )
2
𝑑𝑡
𝑑𝑡
𝑘
𝑚
𝑑2 𝑥
𝑑𝑥
+𝑐
+ 𝑘𝑥 = 𝑧𝑜 𝑘√1 + (2𝑟𝜉)2 sin(𝜔𝑓 𝑡 +∝𝑧 )
2
𝑑𝑡
𝑑𝑡
𝑐𝜔𝑓
∝𝑧 = 𝑡𝑎𝑛−1 (
) = 𝑡𝑎𝑛−1 (2𝑟𝜉)
𝑘
𝑑2 𝑥
𝑑𝑥
This is same as 𝑚 𝑑𝑡 2 + 𝑐 𝑑𝑡 + 𝑘𝑥 = 𝑓𝑜 𝑠𝑖𝑛𝜔𝑓 𝑡 except for the addition of angle ∝ in the sine function
Therefore steady state response
𝑥𝑝 = 𝑧𝑜 √1 + (2𝑟𝜉)2 𝛽 sin(𝜔𝑓 𝑡 +∝𝑧 −∝)
Magnification factor/Transmissibility
𝑥𝑝
√1 + (2𝜉𝑟)2
=
𝑧𝑜 √(1 − 𝑟 2 )2 + (2𝑟𝜉)2
Thus the expression for the relative transmission of the support motion to the oscillator is given by
𝑥𝑝
√1 + (2𝜉𝑟)2
=
× sin(𝜔𝑓 𝑡 +∝𝑧 −∝)
𝑧𝑜 √(1 − 𝑟 2 )2 + (2𝑟𝜉)2
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If r=0, Tr=1.0
If 𝑟 = √2, Tr=1
Transmitted force
In the case of the support motion, the force carried by the support can be obtained
𝑑𝑥 𝑑𝑧
𝐹𝑡 = 𝑐 ( − ) + 𝑘 (𝑥 − 𝑧)
𝑑𝑡 𝑑𝑡
𝐹𝑡 = −𝑚
𝑑2 𝑥
𝑑𝑡 2
𝑥𝑝 = 𝑧𝑜 √1 + (2𝑟𝜉)2 𝛽 sin(𝜔𝑓 𝑡 +∝𝑧 −∝)
𝑑𝑥
= 𝑧𝑜 𝜔𝑓 √1 + (2𝑟𝜉)2 𝛽 cos(𝜔𝑓 𝑡 +∝𝑧 −∝)
𝑑𝑡
𝑑2 𝑥
= −𝑧𝑜 𝜔𝑓 2 √1 + (2𝑟𝜉)2 𝛽 sin(𝜔𝑓 𝑡 +∝𝑧 −∝)
𝑑𝑡 2
Therefore
𝐹𝑡 = 𝑚. 𝑧𝑜 𝜔𝑓 2 √1 + (2𝑟𝜉)2 𝛽 sin(𝜔𝑓 𝑡 +∝𝑧 −∝)
𝑘
𝐹𝑡 = ( 2 ). 𝑧𝑜 𝜔𝑓 2 √1 + (2𝑟𝜉)2 𝛽 sin(𝜔𝑓 𝑡 +∝𝑧 −∝)
𝜔
𝐹𝑡 = 𝑘. 𝑧𝑜 𝑟 2 √1 + (2𝑟𝜉)2 𝛽 sin(𝜔𝑓 𝑡 +∝𝑧 −∝)
10
Mass motion relative to the support.
The relative displacement denoted as z can be written as z=x-y
𝑧=𝑥−𝑦
𝑑𝑧 𝑑𝑥 𝑑𝑦
=
−
𝑑𝑡 𝑑𝑡 𝑑𝑡
𝑑2 𝑧 𝑑2 𝑥 𝑑2 𝑦
=
−
𝑑𝑡 2 𝑑𝑡 2 𝑑𝑡 2
𝑑2 𝑥
𝑑𝑥
Equation 𝑚 𝑑𝑡 2 + 𝑐 𝑑𝑡 + 𝑘𝑥 = 𝑧𝑜 (𝑘𝑠𝑖𝑛𝜔𝑓 𝑡 + 𝑐𝜔𝑓 𝑐𝑜𝑠𝜔𝑓 𝑡) can be written as
𝑚(
𝑑2 𝑧 𝑑2 𝑦
𝑑𝑧
+ 2 ) = −𝑘𝑧 − 𝑐
2
𝑑𝑡
𝑑𝑡
𝑑𝑡
Which can be written as
𝑚
𝑑2 𝑧
𝑑𝑧
𝑑2 𝑦
+
𝑐
+
𝑘𝑧
=
−𝑚
= 𝑚𝑧𝑜 𝜔𝑓 2 𝑠𝑖𝑛𝜔𝑓 𝑡
𝑑𝑡 2
𝑑𝑡
𝑑𝑡 2
Therefore steady state response (particular solution)
𝑥𝑝 =
𝑥𝑝 =
𝑚𝑧𝑜 𝜔𝑓 2 /𝑘
√(1 − 𝑟 2 )2 + (2𝑟𝜉)2
𝑧𝑜 𝑟 2
√(1 − 𝑟 2 )2 + (2𝑟𝜉)2
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× sin(𝜔𝑓 𝑡−∝)
× sin(𝜔𝑓 𝑡−∝)