The length of a simple pendulum is 0.79 m and the mass of the particle (the “bob”) at theend of the cable is 0.24 kg. The pendulum is pulled away from its equilibrium position by anangle of 8.50ᵒ and released from rest. Assume that friction can be neglected and that theresulting oscillatory motion is simple harmonic motion.(a) What is the angular frequency of the motion?(b) Using the position of the bob at its lowest point as the reference level, determine the totalmechanical energy of the pendulum as it swings back and forth ,(c) What is the bob’s speed as it passes through the lowest point of the swing?
Question
The length of a simple pendulum is 0.79 m and the mass of the particle (the “bob”) at theend of the cable is 0.24 kg. The pendulum is pulled away from its equilibrium position by anangle of 8.50ᵒ and released from rest. Assume that friction can be neglected and that theresulting oscillatory motion is simple harmonic motion.(a) What is the angular frequency of the motion?(b) Using the position of the bob at its lowest point as the reference level, determine the totalmechanical energy of the pendulum as it swings back and forth ,(c) What is the bob’s speed as it passes through the lowest point of the swing?
Solution
(a) To find the angular frequency of the motion, we can use the formula:
ω = √(g/L)
where ω is the angular frequency, g is the acceleration due to gravity (approximately 9.8 m/s²), and L is the length of the pendulum (0.79 m).
Plugging in the values, we have:
ω = √(9.8/0.79) ≈ 3.14 rad/s
Therefore, the angular frequency of the motion is approximately 3.14 rad/s.
(b) The total mechanical energy of the pendulum can be determined using the formula:
E = mgh + (1/2)mv²
where E is the total mechanical energy, m is the mass of the bob (0.24 kg), g is the acceleration due to gravity (9.8 m/s²), h is the height of the bob from the reference level (which is the lowest point), and v is the velocity of the bob.
Since the bob is released from rest, its initial velocity is zero. At the lowest point of the swing, the height h is equal to the length of the pendulum L (0.79 m).
Plugging in the values, we have:
E = (0.24 kg)(9.8 m/s²)(0.79 m) + (1/2)(0.24 kg)(0 m/s)² = 1.87 J
Therefore, the total mechanical energy of the pendulum as it swings back and forth is approximately 1.87 J.
(c) The bob's speed as it passes through the lowest point of the swing can be determined using the formula:
v = ω√(L² - h²)
where v is the speed of the bob, ω is the angular frequency (3.14 rad/s), L is the length of the pendulum (0.79 m), and h is the height of the bob from the reference level (which is the lowest point).
At the lowest point of the swing, the height h is equal to zero.
Plugging in the values, we have:
v = (3.14 rad/s)√(0.79 m)² - 0² = 3.14 m/s
Therefore, the bob's speed as it passes through the lowest point of the swing is approximately 3.14 m/s.
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