Question

Three identical masses are coupled together by four identical springs. The position of the left-most mass is 21, the position of the next mass is ry and the final mass is located at position Z3, as shown in the diagram below. பண்டண்டண்டண் | m m m X2 Using Newton's second law, we find the following equations govern the motion of these three masses. _m = =-kz - ke(z) – 22) m" --k(x2 – £1 ) - k:(:2 – £3) m =-k(zy - *2) - k:ry The natural modes of this system occurs for the following combinations of coordinates: ta = 1 – 3 with angular frequency u's T = 11 + 1/232 + 3y with angular frequency ide re = 11 – 1/232 + 3y with angular frequency ide- Given this, we can identify the natural frequencies by comparing our equation to the template simple harmonic oscillator equation: ' If // = 7.8kg and k = 388N/m, what is the smallest natural frequency with which this system will oscillate? =-யா , Answer: 4.9872 x The correct answer is: 5.40 Try another question like this one

Answer 1

The three equation can be written as ma+ku, tk ( 2) -O müz + k (0-2) + K (M,-43)=0 e m 3 + k (x2-12) + k*3 = 0 Normal co-ordinates are kg *W, Ha- ,-" nb = x + 2202 + x3 x = x-a2u2+ "z make Normal co-ordinates I when we are are forme formed linear motion combinations of Equation ahons normal of Equation og co-ordinates should These satisfy the . Linear Combination first combination Xa=%,-3 Subtracting ② from o (mäg- mög) + kx+ K2,- Ishhz – 15x3 +1662 - K mg = 0 m (ä; - g) + k ( 22. - 243] =0 (ä;-žg) = - 2* (*-23) Řa = -2K xa [ from © ] compaining with si a wa 2 ä; z-w?n. - lis La Nm

Әimlаvlа до multiplying мен и+ 2 и 4 з eq. ③ with ars and adding 0 , "С" + 2х, + ) 4 КМ,4 кг, - kж, 12 км, - 2 Kчі 4 км 2- A2 км + км. - К, 4kжа го m (+, + 2, 4,) : «(2-52) », + 2 (2-1)^2 + (2-1)*3]. m( 14 N 24, +7 4 (2-12) 4 (ч- 22+ ) = о х, — (2-42) * again compairing |2-2) А ) D - Fox same x = xp - Az x2 + 3 as above just changing sign of ③ m( я, - 52 + , ) км, км, - Kх, - 2 км, 1 52. км, -52 км, 4 2 км, +k, - Kх, + \{x3 m( * - J2, 45, ) + к 5 (2 + S2 - Г2 (24 п.)2 + (24.Szy13) , – „1, +, + 2+ -52 (ч. – 2 м, м,) го = - C24 A2) сх. o - 24 каде

- * clearly Wb is smallest Wo = (2-2) k -N2) 388 7.8 16 = 0.585 * 388 al - 7.8 i wo = 5.39 radls
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You can find w_{​​​​​​b}​​​​​ to more decimal places. I just rounded of to 2.