Damped free Vibration - Numerical 2

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The disc of a torsional pendulum has a moment of inertia of 600 kgcm2kg \sdot cm^2 and is immersed in viscous fluid. The brass shaft attached to it is of 10 cm10 \ cm diameter and 40 cm40 \ cm long. When the pendulum is vibrating the observed amplitudes on the same side of the rest position for successive cycle are 9°9 \degree, 6°6 \degree, 4°4 \degree. Determine the following:

  1. Logarithmic Decrement
  2. Damping torque at unit velocity
  3. The periodic time of vibration

Assume for the brass shaft, G=4.41010N/m2G = 4.4⨉10^{10} N/m^2. What would be the frequency if the disc is removed from the viscous fluid?

Solution

Given data:

Moment of inertia of disc,  I=600 kgcm2=600×104 kgm2=0.06 kgm2Diameter of the shaft,  d=10 cm=0.1 mLength of the shaft,  L=40 cm=0.4 mInitial amplitude,  x0=9°Second amplitude,  x1=6°Third amplitude,  x2=4°\begin{aligned} \text{Moment of inertia of disc,} \ \ I &= 600 \ kg \sdot cm^2 \\ &= 600 \times 10^{-4} \ kg \sdot m^2 \\ &= 0.06 \ kg \sdot m^2 \\ \\ \text{Diameter of the shaft,} \ \ d &= 10 \ cm = 0.1 \ m \\ \\ \text{Length of the shaft,} \ \ L &= 40 \ cm = 0.4 \ m \\ \\ \text{Initial amplitude,} \ \ x_0 &= 9 \degree \\ \\ \text{Second amplitude,} \ \ x_1 &= 6 \degree \\ \\ \text{Third amplitude,} \ \ x_2 &= 4 \degree \\ \end{aligned}

1. The logarithmic decrement:

δ=loge(x0x1)=loge(96)=loge(1.5)=0.405\begin{aligned} \delta &= {\log}_e \bigg( {x_0 \over x_1}\bigg) \\ \\ &= {\log}_e \bigg( {9 \over 6}\bigg) \\ \\ &= {\log}_e (1.5) \\ \\ &= 0.405 \end{aligned}

fig_damped-free-vibration-numerical-1

2. Damping torque at unit velocity:

  • The damping factor is,
ξ=δ4π2+δ2=0.4054π2+(0.405)2=0.0642\begin{aligned} \xi &= {\delta \over \sqrt{4 {\pi}^2 + {\delta}^2}} \\ &= {0.405 \over \sqrt{4 {\pi}^2 + (0.405)^2}} \\ &= 0.0642 \end{aligned}
  • The torsional stiffness of shaft is,
Kt=GJL=Gπ32(d)4L=4.4×1010×π32(0.1)40.4=1.07×106 Nm/rad\begin{aligned} K_t &= {GJ \over L} \\ \\ &= {G \sdot {\pi \over 32} \sdot (d)^4 \over L} \\ \\ &= {4.4 \times 10^{10} \times {\pi \over 32} \sdot (0.1)^4 \over 0.4} \\ \\ &= 1.07 \times 10^6 \ Nm/rad \end{aligned}
  • The natural circular frequency of system is,
ωn=KtI=1.07×1060.06=4242.5  rad/s\begin{aligned} {\omega}_n &= \sqrt{K_t \over I} \\ \\ &= \sqrt{1.07 \times 10^6 \over 0.06} \\ \\ &= 4242.5 \ \ rad/s \end{aligned}

Now, Damping torque,

Ft=ctx˚F_t = c_t \sdot \mathring{x}

\therefore Damping torque at unit velocity, Ftx˚=ct{F_t \over \mathring{x}} = c_t


Now, we know that;

ξ=ctcc\xi = {c_t \over c_c} ct=ξcc=ξ(2Iωn)=0.0642(2×0.06×4242.5)=32.74  Nms/rad\begin{aligned} \therefore c_t &= \xi \sdot c_c \\ &= \xi (2 \sdot I \sdot {\omega}_n) \\ &= 0.0642(2 \times 0.06 \times 4242.5) \\ &= 32.74 \ \ N \sdot m \sdot s / rad \end{aligned}

3. The time period of damped vibration:

tp=2πωd=2πωn1ξ2=2π(4242.5)1(0.0642)2=0.00148  second\begin{aligned} t_p &= {2\pi \over {\omega}_d} \\ \\ &= {2\pi \over {\omega}_n \sqrt{1 - {\xi}^2}} \\ \\ &= {2\pi \over (4242.5) \sqrt{1 - (0.0642)^2}} \\ \\ &= 0.00148 \ \ second \end{aligned}

4. Frequency, if disc is removed from viscous fluid:

fn=ωn2π=4242.52π=675.21 Hz\begin{aligned} f_n &= {{\omega}_n \over 2\pi} \\ \\ &= {4242.5 \over 2\pi} \\ \\ &= 675.21 \ Hz \end{aligned}

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