Question

Problem 5: The spring-mass system shown has spring constants ky = 24 kN/m and kz = 36 kN/m with a suspended mass of 35 kg at A. If the block is displaced 50 mm below its equilibrium position and released with no initial velocity, determine: a) The circular natural frequency, the natural frequency, and the period b) The position, velocity, and acceleration of the block after a time of 30 seconds k2 mm ki A

Answer 1

Guiren: m = 35kg WWW K2 KI K = 24 kN/m K₂ = 36 kN/m. А to we can aus cosider this system equivalent simple Apring block system follow. with Apring Constant (keq) As we sed the spoings k w K2 connected in parallel Connection. arre Kega keq = KitKz 24+36 A Kega 60 KN/m Now as we know & circular natural frequery Kear 60 35 lo 30 vad S. 0.208 rad & Natural frequency 27 m - 4.79 Seconds. & Time period ,T = 20 Keq

Drawing phasor diagram of equiralent Spring block Afsta. AY. Amplitude of 1 Jo-wit BX oscilation ele goutation from equillabrium position 50mm = 50x1013 m. A. = 50 mm, It=0s. y= = A sin (wt +$). when A= amplitude y=position of mass. W= natural feaven d = phase measured position. from ④ x aus. I t= time in seconds - t from mean y = 2 50x10² sim (1:30€ cos (1.30t). -3 - 50x10 OG to, At At y = -50X10-3. 38x 10-3 mm ans t=305 & dy at t=305 VE+0.0516 m/s position at t=30s. Differiting y gives velocity, =V=+1.30X50X10 3 cos (1.30t - 7/2). dt velocity, acceleration a= W = (1.30x50 x 10-3 sen (1.304-262) m/s2 negative Aign meas 2 a =-0.051 m2 acche att=30s. aan towards mean position

Guiren: m = 35kg WWW K2 KI K = 24 kN/m K₂ = 36 kN/m. А to we can aus cosider this system equivalent simple Apring block system follow. with Apring Constant (keq) As we sed the spoings k w K2 connected in parallel Connection. arre Kega keq = KitKz 24+36 A Kega 60 KN/m Now as we know & circular natural frequery Kear 60 35 lo 30 vad S. 0.208 rad & Natural frequency 27 m - 4.79 Seconds. & Time period ,T = 20 Keq