Question

A simple pendulum consisting of a bob attached to a cord oscillates in a vertical plane with a period of 1.3 s. Assuming simple harmonic motion and knowing that the maximum velocity of the bob is 0.4 m/s, determine (a) the amplitude of the motion in degrees, (b) the maximum

Answer 1

a) = 11.29°

b) = 1.93 m/s²

Step-by-step explanation:

θ = θ(m) sin(w(n).t + Φ)

w(n) = 2π / t

w(n) = 6.284 / 1.3

w(n) = 4.834 rad/s

θ• = θ(m).w(n).cos (w(n).t + Φ)

θ•(m) = θ(m).w(n)

v(m) = L.θ•(m)

v(m) = L.θ(m).w(n)

θ(m) = v(m) / L.w(n)

For a simple pendulum,

w(n) = √(g / l)

l = g / w(n)²

l = 9.8 / 4.834²

l = 9.8 / 23.368

l = 0.42 m

Remember, θ(m) = v(m) / l.w(n)

θ(m) = 0.4 / (0.42 * 4.834)

θ(m) = 0.4 / 2.03

θ(m) = 0.197 rad

Converting to °, we have

0.197 * 360 * 1/2π

0.197 rad = 11.29°

Thus, the amplitude of the motion is 11.29°

a(t) = l.θ(t)

Tangential acceleration occurs when θ(t) is maximum, so

θ(t) = -θ(m).w(n)².sin(w(n)t + Φ)

θ(t max) = θ(m).w(n)²

a(t max) = l.θ(m).w(n)²

a(t max) = 0.42 * 0.197 * 4.834²

a(t max) = 1.93 m/s²