The spring constant k is 17747N/m. The position of oscillator when t=0 S is 0.43 m_. The time(t>0 s)the oscillator first at maximum distance from equilibrium is_0.44 S_. The maximum speed of the oscillator is_ 9.42 m/s_. The magnitude of the maximum acceleration of the oscillation is 17747 m/s2 Solution From the equation x(o)=(0.50m)cos[(. 0T rad/s)t+r rad 6.0 (a)A=0.5m (b)o=67 rad/s (c)v=2_6z 3HZ 2x2丌 6丌3 →k=a2m=(6x)2×05=182=17747N/n (When (=0 x=0.5cos =0433m g)=105co(60×/0→c06m+a)=1=67+x=kz k丌-2)k=0,±1,±2 Ifk=1t=-丌=0.44s (h)x=0.5c0s(6丌1+-) V=,=-3rsin(6t+-) v=3丌=942m/s )a==-18xc06+6The spring constant k is 177.47 N/m . The position of oscillator when t = 0 s is 0.43 m . The time (t > 0 s ) the oscillator first at maximum distance from equilibrium is 0.44 s . The maximum speed of the oscillator is 9.42 m/s . The magnitude of the maximum acceleration of the oscillation is 177.47 m/s2 . Solution: From the equation ] 6.0 rad ( ) (0.50m) cos[(6.0π rad/s) π x t = t + , we get (a) A = 0.5m (b) ω = 6π rad/s (c) 3 Hz 2 6 2 = = = π π π ω ν (d) s 3 1 6 2 2 = = = π π ω π T (e) (6 ) 0.5 18 177.47 N/m 2 2 2 ω = ⇒ k = ω m = π × = π = m k (f) When t = 0 0.433 m 4 3 6 = 0.5cos = = π x g) π π π π π π x = πt + = ⇒ t + = ± ⇒ t + = k 6 ) 1 6 6 ) 0.5 cos(6 6 0.5cos(6 k t s t k k 0.44 36 5 If 1 ) 0, 1, 2, 6 ( 6 1 = = = ⇒ = − = ± ± π π π π K (h) ) 6 0.5cos(6 π x = π t + ) 6 3 sin(6 d d π = = − π π t + t x v 3 9.42 m/s vmax = π = (i) ) 6 18 cos(6 d d 2 π = = − π π + t v a