The length of a simple pendulum is 0.65 m and the mass of the particle (the Solu

Question

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

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.