Question

The length of a simple pendulum is 0.79 m and the mass of the particle (the

Answer 1

SHM equ >> y = a sin wt = a sin (p)
find v (=dy/dt), then acceleration (f=dv/dt) you get
f = - w^2 y
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where w = 2pi f = 2pi/T is the angular frequency
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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
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pendulum T = 2pi sqrt(L/g)
w=2pi/T = sqrt(g/L)
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w^2 = g/L = 9.8/0.79
w = 3.52 rad/sec
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from above equ, one can show that
v^2 = w^2 (a^2 - y^2)
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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
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PE = spring = 0.5 ky^2
k = m w^2
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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
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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