A mass of 85 kg is supported on a spring which deflects 18 mm under the weight of the mass. The vibrations of the mass are constrained to be linear and vertical. A dashpot is provided which reduces the amplitude to one-quarter of its initial value in two complete oscillations. Calculate magnitude of the damping force at unit velocity and periodic time of damped vibration.
Solution:
Fig-1
Given data:
mass m = 85 \ m = 85 m = 85 kg
Static deflection, x = 18 mm = 0.018 m \ x = 18\ \text{mm} = 0.018\ \text{m} x = 18 mm = 0.018 m
Let,
x 0 = x_0 = x 0 = initial amplitude
x 2 = x_2= x 2 = final amplitude after two complete cycle = 1 4 = {1\over4} = 4 1
Taking ratio,
x 0 x 2 = x 0 1 4 x 0 = 4 {x_0 \over x_2}={x_0 \over {1 \over 4} x_0}=4 x 2 x 0 = 4 1 x 0 x 0 = 4
Damping force:
F = c . x ˚ = c \text F = c.\mathring x= c F = c . x ˚ = c
Now,
ξ = c c c OR ξ = c 2 m ω 0 ( as c c = 2 m ω n ) ∴ c = ( 2 m ω n ) ξ … … … … ( 1 ) \begin{aligned}
\xi= {c \over c_c} \quad \text {OR} \quad &\xi= {c \over 2m\omega_0} \quad (\text {as} \quad c_c=2m\omega_n)\\
\therefore \ &c=(2m\omega_n)\xi \ \dots\dots\dots\dots \ (1)
\end{aligned} ξ = c c c OR ∴ ξ = 2 m ω 0 c ( as c c = 2 m ω n ) c = ( 2 m ω n ) ξ ………… ( 1 )
1. The Logarithmic decrement:
δ = 1 n l o g e ( x 0 x n ) = 1 2 l o g e ( x 0 x 2 ) = 1 2 l o g e ( 4 ) \begin{aligned}
\delta &= {1 \over n} log_e\bigg ({x_0 \over x_n}\bigg)\\
&= {1 \over 2} log_e\bigg ({x_0 \over x_2}\bigg)\\
&= {1 \over 2} log_e(4)
\end{aligned} δ = n 1 l o g e ( x n x 0 ) = 2 1 l o g e ( x 2 x 0 ) = 2 1 l o g e ( 4 )
2. Damping factor:
ξ = δ 4 π 2 + δ 2 = δ 4 π 2 + δ 2 = 0.1091 \begin{aligned}
\xi &= {\delta \over \sqrt{4 \pi^2 + {\delta}^2} } \\
&={\delta \over \sqrt{4 \pi^2 + {\delta}^2} }\\
&= 0.1091
\end{aligned} ξ = 4 π 2 + δ 2 δ = 4 π 2 + δ 2 δ = 0.1091
3. Frequency of undamped free vibration:
The natural circular frequency of vibration is,
ω n = k m = m g x m = g x = 9.81 0.018 = 23.34 rad/s \begin{aligned}
\omega_n &= {\sqrt{k \over \ m}} \\
&= {\sqrt{m \ g \over \ x \ m}}\\
&= {\sqrt{g \over x}} \\
&= {\sqrt{9.81 \over 0.018}}\\
&= 23.34 \ \text {rad/s}
\end{aligned} ω n = m k = x m m g = x g = 0.018 9.81 = 23.34 rad/s
From equation (1); we get,
Damping force at unit velocity;
F = c = ( 2 m ω n ) ξ = ( 2 x 85 x 23.34 ) x ( 0.1091 ) = 91.67 N \begin{aligned}
\text F = c &= (2m\omega_n)\xi \\
&= (2 \ \text x \ 85\ \text x \ 23.34)\ \text x \ (0.1091) \\
&= 91.67 \ \text N
\end{aligned} F = c = ( 2 m ω n ) ξ = ( 2 x 85 x 23.34 ) x ( 0.1091 ) = 91.67 N
The time period of damped vibration is,
t p = 2 π ω d = 2 π ω n 1 − ξ 2 = 2 π ( 23.34 ) 1 − ( 0.1091 ) 2 = 0.2708 sec \begin{aligned}
t_p &= {2\pi \over \omega_d} \\
&= {2\pi \over \omega_n \ \sqrt {1 -\xi^2}} \\
&= {2\pi \over (23.34) \ \sqrt {1 -(0.1091)^2}} \\
&=0.2708 \ \text {sec}
\end{aligned} t p = ω d 2 π = ω n 1 − ξ 2 2 π = ( 23.34 ) 1 − ( 0.1091 ) 2 2 π = 0.2708 sec
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