5.2. A rigid bar of mass per unit length m carries a point mass M at its right end. The bar is supported by two springs, as shown in Fig. 5.21. Derive the differential equations for the translation and rotation of the mass center. Assume small motions. let k₁= k, k₂ = 2k, M = mL, and calculate the natural frequencies and natural modes. Plot the modes. A.F(t) k₁ m |— ¾1 A www L +²4 L k₂ FIGURE 5.21 Mass supported by springs through a rigid bar M

Elements Of Electromagnetics
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5.2. A rigid bar of mass per unit length m carries a point mass M at its right end. The bar is
supported by two springs, as shown in Fig. 5.21. Derive the differential equations for the
translation and rotation of the mass center. Assume small motions.
let k = k, ky = 2k, M = mL, and calculate the natural
frequencies and natural modes. Plot the modes.
A. F(t)
ww
k₁
m
| – 31 L
A
www
11
+²4
L
S
FIGURE 5.21
Mass supported by springs through a rigid bar
M
Transcribed Image Text:5.2. A rigid bar of mass per unit length m carries a point mass M at its right end. The bar is supported by two springs, as shown in Fig. 5.21. Derive the differential equations for the translation and rotation of the mass center. Assume small motions. let k = k, ky = 2k, M = mL, and calculate the natural frequencies and natural modes. Plot the modes. A. F(t) ww k₁ m | – 31 L A www 11 +²4 L S FIGURE 5.21 Mass supported by springs through a rigid bar M
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