A 250-gram mass is connected to a spring (k = 81 N/m). The mass is pulled out 15

Question

A 250-gram mass is connected to a spring (k = 81 N/m). The mass is pulled out 15 cm from its equilibrium position, and released at t = 0. During the subsequent oscillations, the position of the mass is described by x(t) = xm cos (wt). (1) What is xm? (2) What is w? (3) What is its position s at t = 1.00 s? (4) What is vx at t= 1.00 s? (5) What is its maximum speed? (6) What is its maximum kinetic energy? (7) What is its maximum potential energy? (8) What is the system?s mechanical energy (E = K + U)? (9) What is its speed at x = 5 cm? Hint: K(x) = E- U(s). (10) How many oscillations will this mass complete in one minute?

Explanation / Answer

1) xm = 15 cm = 0.15m

2) w = sqrt(k/m) = sqrt(81/0.25) = 18 rad/s

3) x = 0.15*cos(18*1) = 0.099 m = 9.9 cm

4) vx = -xm*w*sin(wt) = -0.15*18*sin(18*1) = 2.027 m/s

5) vmax = xm*w = 2.7 m/s

6) Kmax = 0.5*m*vmax^2 = 0.5*0.25*2.7^2 = 0.91125 J

7) Potential enrgy Umax = 0.5*K*A^2 = 0.5*81*0.15^2 = 0.91125 J

8) E = KEmax = Umax = 0.91125 J

9) v = w*sqrt(A^2-x^2) = 18*sqrt(0.15^2-0.05^2) = 2.54 m/s

10) frequency f = w/(2*pi) = 18/(2*3.142) = 2.86 oscillations per second

per minute it will be 2.86*60 = 171.6 oscillations

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