Answer:
a) = 11.29°
b) = 1.93 m/s²
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
Step-by-step explanation: