Question

-NM m 1 a H x. I m the two system configurations shown above. The second configuration, on the right, the natural frequency, in Hz, is measured at %25 more than the frequency of the first configuration, on the left. Find, k2, stiffness coefficient; if k, = 1 N/m.

Answer 1

Solution criven data : Ki= 1 N/m Wn = 1.25 Why Right side Left side žkz Here: K = Shiffness Coefficient Wn, = natural frequency for Fig. O Why = naturat frequency for fig. ③ Well know that Natural frequency Wy v kreeg Here keq e equivalent Stipfosess coefficient Meg = Equivalent mass Par fig ① keq, = ki + (1/kt . Keg, = ky+ Kxkz - Series part Kitko for Fig 2 keq = Kitk₂ .K&K are in parallel Now Why = 1.25 Wn, from giver Req2 31. 25 Keep mear

Squaring Both side On Keyz = 1.5625 meqz keegi megy i Kega = 1.5625 key, (.meg, = Mega = m) mass is Same) Ritka = 1.5625 (kit Kixk2 ) mas kitk Put value of ki = 1 N/m 1+ k, = 1.5628 [1 + 127 on solving | K₂=1.5 N/m & Stiffness Coefficient K₂ is 1.5 N/m