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It gives us a value of approximately 2.87 meters. This is the position of the 1.2-kg slider that results in a period of 1 s.
To determine the position of the 1.2-kg slider that results in a period of 1 s, we need to first understand the physics of the situation. A uniform slender rod is a type of simple harmonic oscillator, which means that it will oscillate back and forth around its equilibrium position with a certain period. The period of a simple harmonic oscillator is determined by its mass and the stiffness of the force that is restoring it to its equilibrium position. In the case of a uniform slender rod, the restoring force is provided by the rod's own weight.
The period of a simple harmonic oscillator is given by the formula:
[tex]T = 2 * \pi * \sqrt{\frac{m}{k} }[/tex]
Where T is the period, m is the mass of the object, and k is the stiffness of the force that is restoring it to its equilibrium position. In the case of a uniform slender rod, the mass is 3 kg and the stiffness is the force of gravity, which is 9.8 Newtons/kg.
Plugging these values into the formula above, we get:
[tex]T = 2 * \pi * \sqrt{\frac{3}{9.8} }[/tex]
This simplifies to:
[tex]T = 2 * \pi * \sqrt{(0.306)}[/tex]
Which gives us a period of approximately 0.9 seconds.
To determine the position of the 1.2-kg slider that results in a period of 1 s, we need to solve the equation:
[tex]T = 2 * \pi * \sqrt{\frac{m}{k} } = 1 s[/tex]
for the position x. Since we know the mass of the slider and the stiffness of the restoring force, we can simply plug in these values and solve for x.
The mass of the slider is 1.2 kg, and the stiffness of the restoring force is 9.8 N/kg, so we have:
[tex]1 s = 2 * \pi * \sqrt{\frac{1.2}{9.8} }[/tex]
Solving for x, we get:
[tex]x = \sqrt{\frac{9.8}{1.2} }[/tex]
This simplifies to:
x = [tex]\sqrt{8.17}[/tex]
x = 2.87 meters.
Learn more about oscillations, here https://brainly.com/question/28994371
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