The length of a simple pendulum is 0.79 m and the mass at the end of the spring

Question

The length of a simple pendulum is 0.79 m and the mass at the end of the spring is 0.24 kg. The pendulum is pulled away from its equilibrium position by an angle of 8.50 degrees and released from rest. Neglecting friction and assuming SHM, what is the speed of the mass as it passes through the lowest point of the swing?

Explanation / Answer

SHM equ >> y = a sin wt = a sin (p) find v (=dy/dt), then acceleration (f=dv/dt) you get f = - w^2 y --------- where w = 2pi f = 2pi/T is the angular frequency ------------------------ restoring force = - mg sin p acceleration = - g sin p now g sin p = w^2 y here sin p = y/L (small angle) g*y/L = w^2 y --------------------------------- pendulum T = 2pi sqrt(L/g) w=2pi/T = sqrt(g/L) -------------------------- w^2 = g/L = 9.8/0.79 w = 3.52 rad/sec --------------------------------- from above equ, one can show that v^2 = w^2 (a^2 - y^2) ---------------------- KE of bob = 0.5 mv^2 = 0.5 m w^2 (a^2 - y^2) PE of bob = 0.5 mw^2 * y^2 --------------------------------- PE = spring = 0.5 ky^2 k = m w^2 -------------------- ME = KE+PE = 0.5 mw^2 a^2 where a = amplitude = L sin p = 0.79* sin 8.5 ME = 0.5*0.24*(3.52)^2 * (0.79 sin 8.5)^2 ME = 0.020 Joule -------------------- speed at lowest point is maximum, and when y =0 v^2 = w^2 (a^2 - y^2) v^2(mid) = (aw)^2 v(mid) = (aw) = 0.79 sin 8.5 * 3.52 = 0.411 m/s

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