Question

3. A body weighing 25 lb is suspended from a spring of constant k=160 lb/ft. The system is shown in Figure 1. At time t=0, it has a downward velocity of 2 ft/sec as it passes through the position of static equilibrium. Determine: (a) the static spring deflection. (b) the natural frequency of the system in both rad/sec and cycles/sec. k = 160 lb/ft (c) the system period (d) the displacement x as a function of time, where x is measured from the position of static equilibrium W = 25 lb (e) the maximum velocity attained by the mass (f) the maximum acceleration attained by the mass. Fig. 1

Answer 1

ye no 25 lb W2 gb eb Kz 160 lb/ft #20 ; x = 2 ft/sec &k=160 lb/ft - w225 eb Static I deflections alo attaching Fouilibrium @ Speving forces weight [ unstreched position Is ak 8 spring 153 of from popu FBD blook; Pg - Wao F. B. D. Ps 2 W] 32 Total of muss Station 1.8 20.15625 ft or 8 ft By Rayleigh's methods Natural Angular frequency on a sa static deflection ene 2234 +/secs wna 32 zaal Natural frequency radh in I. ne 14.3108 haars & frequency in cycles/; fn a con [ Wm2 et fr] fm2 3483 Se 529 ft l. fn 2 9.2776 Cycles/sed Natural frequency Hz in Cycles/

9 C system time period; lobos TT 2 Second - Cycle period 1. 0.439 Time second @ suppose displacement of block at any position is given by; TOCA X sin cont oL | velocity; az xwn cos (wnto Acceleration; t x an sin contro) Now's at equilibrium position [#20]; starting of st Vibratory simple Harmonic motion : velocity sc (0) = 2 ft/sec (taking down word velocity tve] (sign Convention such M ats VW J.Ves 1 At some] instant z Putting sc (0) 2a fall sec in ☆ (4) 2x un cos (Wnt - 0) where X2 Amplitude of xlox²4 - [w]. motion x Equilibrium de phase difference Position Both are {x,o] arbitary constant & can be fou fou found by given Initial & Boundary condition. oc (0) 2 x Wn cos [wnxo -o] 2 + x(0) 2x un cost & 2 x * 326 (03

S52 x 6030 & since; at equilibrium position, 21 pm we refrence the motion, displacement of Block can be taken zero. so; at t20 ; x(0) = 0 {Equilibrium Position ? Pretting this condition in x (+) 2 X sin (wat-o 21022 x sin [Wnxo -$] - since X2 Amplitude & for vibration Amplitude can - not be zero in any circumstance soj sin {-0] =0 -sinc0720 sol 102011 Phuse Difference so; from enn ou x cospa 5/6 xloso a S/16 x (1)2 56/16 Amplitude 1x20. 1398 ft or & ft - Nowo's soi egn of motion is > *(+)2 sin [ t] : ft| - Displacement as function of time

xa xwn cos [wnto] I gi(t) = q los [ 32 + a velocity as function of time. when to - given function is mascimum cos 3720 here nzo,1,2,... zt a NT J5 occurs max. velocity n T 55 s t2 t2 Let us, take no o tros) - se (0) 22 ft/s Masumum value we 1" we. maximum get velocity. [As "cos function" having maximum can repluce "cos function" by "f" value of given function.] öc a X on sin (unt -Ø) že s VBM (2) sim Cepat-po) Acceleration as a za sin ( 34 ) |– A function of time

part can function ① ; occur has when - Similarly, as explained in maximum value of given function sing a sim ( 32 t) 21 since sine maximum value equals to y. so; sin ( 32 +) 21 ,M20,1,2,- 34+2 (anti) I t 2 (anti- #ta Cantil) mo 0,1,2,---| soi musimum acceleration; Costache ämux 2 64 - äemuse 2 64 2 28-6218 /s2 maximum acceleration