Question

The length of a simple pendulum is 0.78 m and the mass of the particle (the "bob") at the end of the cable is 0.26 kg. The pendulum is pulled away from its equilibrium position by an angle of 8.70° and released from rest. Assume that friction can be neglected and that the resulting oscillatory motion is simple harmonic motion.

(a) What is the angular frequency of the motion?
___rad/s

(b) Using the position of the bob at its lowest point as the reference level, determine the total mechanical energy of the pendulum as it swings back and forth.
___J

(c) What is the bob's speed as it passes through the lowest point of the swing?
____m/s

Answer 1

(a) The expression for the simple harmonic motion is -

y = a sin wt = a sin (p)

So, v (=dy/dt), then acceleration (f=dv/dt) you get
f = - w^2 y

where w = 2pi f = 2pi/T is the angular frequency

again we have -

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)

So -

w^2 = g/L = 9.8/0.78 = 12.56

=> w = 3.54 rad/sec

(b) from above equ, we 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 = 0.78* sin 8.7
ME = 0.5*0.26*(3.54)^2 * (0.78 sin 8.7)^2
=> ME = 1.629 *0.0139 = 0.02267 J = 22.7 mJ

(c) 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) = 0.78 sin 8.7 * 3.54 = 0.418 m/s

Therefore, bob's speed as it passes through the lowest point of the swing = 0.418 m/s.