A simple pendulum consists of a 900cm string with a 300kg ma

A simple pendulum consists of a 90.0-cm string with a 3.00-kg mass at the end of it. The pendulum is initially displaced 16.0° from equilibrium and released from rest. Determine the period, horizontal amplitude (in the x-direction), and maximum speed of the pendulum. [1.90e+000 s, 2.48e–001 m, 8.19e–001 m/s]

Solution

Here ,

L = 90 cm = 0.90 m

mass ,m = 3 Kg

theta = 16 degree

for the period ,

period of pendulum = 2pi * sqrt(L/g)

period of pendulum = 2pi * sqrt(0.90/9.8)

period of pendulum = 1.90 s

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Now , for the horizontal amplitude

horizontal amplitude = L * sin(theta)

horizontal amplitude = 0.90 * sin(16)

horizontal amplitude = 0.248 m

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maximum speed of the pendulum is v

0.5 * m * v^2 = m * g * L * (1 - cos(theta))

0.5 * v^2 = 9.8 * 0.90 * (1 - cos(16))

solving for v

v = 0.819 m/s

the maximum speed of pendulum is 0.819 m/s