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A simple pendulum has mass 1.20 $\mathrm{kg}$ and length 0.700 $\mathrm{m} .$ (a) What is the period of the pendulum near the surface of Earth? (b) If the same mass is attached to a spring, what spring constant would result in the period of motion found in part (a)?

a. $T _ { 1 } = 1 \cdot 68 \mathrm { sec }$b.$k = 16.8 \mathrm { N } / \mathrm { m }$

Physics 101 Mechanics

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Chapter 13

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In physics, electromagnetic radiation (EM radiation or EMR) refers to the waves (or their quanta, photons) of the electromagnetic field, propagating (radiating) through space, carrying electromagnetic radiant energy. It includes radio waves, microwaves, infrared, (visible) light, ultraviolet, X-rays, and gamma rays. Electromagnetic waves of different frequency are called by different names since they have different sources and effects on matter. In order of increasing frequency and decreasing wavelength these are: radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays and gamma rays.

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so here for party. We can say that the period of a simple pendulum T would be willing to pie multiplied by the square root of the length divided by the acceleration due to gravity. And so the period of the first system t someone would be equaling two pi multiplied by the square root of point 700 meters. This would be multiple brother divided by 9.80 meters per second squared and this is giving us 1.68 seconds. This would be the end of the period, rather, for the first system, our answer to part way and then for part B. We can't say that the object of this, the period rather oven object spring system would be t equaling two pi multiplied by the square root of the mass divided by K this spring constant and so we know that the period of the second system equals the period of the first system. However, we can then say that to pie multiplied by the square root of the mass. Divided by K, the spring constant equals two pi multiplied by the square root of the length divided by G, the acceleration due to gravity. And so this spring constant K would be equaling the MG divided by the length l and so 1.20 kilograms multiplied by 9.80 meters per second squared. This would be divided by the length of 0.700 meters and we find that the spring constant k ISS 16.8 newtons per meter. This would be our answer for the spring constant for the second system for part B. That is the end of the solution. Thank you for watching.

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