When a 1.3kg mass was attached to a spring, the spring stretched 19cm from its e

Question

When a 1.3kg mass was attached to a spring, the spring stretched 19cm from its equilibrium position. The mass spring system was then set into oscillatory motion by providing the mass with an initial velocity. Its oscillatory motion is then described by the following equation: Y(t) = (0.09m) sin(2pi t/T + pi/5).

(a) find the maximum speed (in m/s)

(b) maximum acceleration of the mass (in m/s2),

(c) Find the position of the mass at t = 0 s in meters.

(d) Find the speed of the mass at t = 0 s in meters

(e) Find the mechanical energy of the system relative to the equilibrium position at t = 0 s.

Explanation / Answer

K= 1.3 * 9.81/ .19 =67.12 N/m

w =sqrt(k/m) =51.53 rad/sec

A=0.09 meter

(a) max veklocity = Aw = 4.65 m/s

(b) max accelration= Aw2 = 0.42 m/s2

(c) Y = A sin (wt+ pi/5)   

at t = 0

Y = 0.09 sin( 51.53* 0 + pi/5 ) = 0.0827 meter

(d) V = Aw cos ( wt + pi/5 )   

at t =0

V =Aw cos ( pi/5) = -1.82 m/s

(e) mch energy = 1/2 mV2     =1/2 *1.3 *1.822 = 2.15 J

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