For small angles, does the pendulum's period of oscillation depend on initial angular displacement from equilibrium? Explain.
1 answer:
Answer:
No, the pendulum's period of oscillation does not depend on initial angular displacement.
Explanation:
Given that,
For small angle, the pendulum's period of oscillation depend on initial angular displacement from equilibrium.
We know that,
The time period of pendulum is defined as
Where, l = length of pendulum
g = acceleration due to gravity
So, The time period of pendulum depends on the length of pendulum and acceleration due to gravity.
It does not depend on the initial angular displacement.
Hence, No, the pendulum's period of oscillation does not depend on initial angular displacement.
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The center of mass of the two objects is the average position of the parts of the two object system
The center of mass of block <em>A</em>, and block <em>B</em> after displacement of block <em>B</em> is at <u>20 m from block </u><u><em>A</em></u>
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Reason:
The given parameters are;
The position of block A and block B at t = 0 is x = 0
The mass of block A, m₁ = 8 kg
Mass of block B, m₂ = 16 kg
Speed of block <em>A</em> = 0 m/s
Speed of block <em>B</em>, v₂ = 10 m/s
Location of the center of mass of the two object at t = 3 s; Required
Solution;
The location of block <em>A</em>, after 3 s is x₁ = 0 (block A is at rest)
The location of block <em>B</em>, = v₂ × t
The location of block <em>B</em>, after 3 s is x₂ = 10 m/s × 3 s = 30 m
The center of mass of two masses are given as follows;
The center of mass of the two objects is at at the position x = <u>20 m</u> (from block <em>A</em>)
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