[amp_mcq option1=”1.00 m\/sec” option2=”1.57 m\/sec” option3=”3.14 m\/sec” option4=”6.28 m\/sec” correct=”option1″]<\/p>\n
The correct answer is $\\boxed{\\text{B. }1.57\\text{ m\/s}}$.<\/p>\n
The maximum velocity of a particle in simple harmonic motion is reached when the particle is at its extreme position. In this case, the particle is 0.5 m from the equilibrium position, and its velocity is given by:<\/p>\n
$$v = \\omega\\sqrt{A^2 – x^2}$$<\/p>\n
where $\\omega$ is the angular frequency of the motion, $A$ is the amplitude of the motion, and $x$ is the displacement of the particle from the equilibrium position.<\/p>\n
We are given that the time period of the motion is 1 s, so the angular frequency is:<\/p>\n
$$\\omega = \\frac{2\\pi}{T} = \\frac{2\\pi}{1\\text{ s}} = 2\\pi\\text{ rad\/s}$$<\/p>\n
We are also given that the amplitude of the motion is 1 m, so the maximum velocity of the particle is:<\/p>\n
$$v = \\omega\\sqrt{A^2 – x^2} = 2\\pi\\sqrt{1^2 – 0.5^2} = 2\\pi\\sqrt{0.75} = 1.57\\text{ m\/s}$$<\/p>\n
The other options are incorrect because they do not represent the maximum velocity of the particle in simple harmonic motion.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”1.00 m\/sec” option2=”1.57 m\/sec” option3=”3.14 m\/sec” option4=”6.28 m\/sec” correct=”option1″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[645],"tags":[],"class_list":["post-6768","post","type-post","status-publish","format-standard","hentry","category-applied-mechanics-and-graphic-statics","no-featured-image-padding"],"yoast_head":"\n