A block whose mass m is 680 kg is fastened to a spring whosespring constant k is

Question

1. A block whose mass m is 680 kg is fastened to a spring whosespring constant k is 65 N/m. The block is pulled a distance x = 11cm from its equilibrium position at x = 0 on a frictionless surfaceand released from rest at t = 0. (a) What is the angular frequency,the frequency, and the period of the resulting motion? (b) What isthe amplitude of the oscillation? (c) What is the maximum speed ofthe oscillating block, and where is the block when it has thisspeed? (d) What is the magnitude of the maximum acceleration of theblock? (e) What is the phase constant for this motion? (f) What isthe displacement function x(t) for this spring-block system?

Explanation / Answer

k=65N/m. m=680kg. a) as angular frequency =k/m=(65/680)=0.309rad/s. frequency f=/2=0.309/2*3.14=0.049 s-1. time period T=1/f=1*0.049=2.04 s. b) as maximum displacement is x=11cm, amplitude A=11cm. c) also v=(A2-x2). as velocity is maximum at x=0; vmax=A=2.04*0.11=0.2244 m/s. d) acceleration is maximum at x=A; amax.=2A.=0.457m/s2. e) as the motion is started from its maximum function is cos so lt theequation of the motion is : x(t)=Acos(t+)                                    (where is the phase constant). as at t=0, the body is at +ve A; A=Acos(0+); cos=1; or =/2. f) so the equation of the function x(t) isx(t)=0.11cos(0.309t+/2).

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