{"id":36033,"date":"2025-06-25T01:40:13","date_gmt":"2025-06-25T01:40:13","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=36033"},"modified":"2025-06-25T01:40:15","modified_gmt":"2025-06-25T01:40:15","slug":"a-particle-undergoing-simple-harmonic-motion-has-a-velocity","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/a-particle-undergoing-simple-harmonic-motion-has-a-velocity\/","title":{"rendered":"A particle undergoing simple harmonic motion has a velocity"},"content":{"rendered":"\n<p>A particle undergoing simple harmonic motion has a velocity 1, when the displacement is X1 anda velocity %, when the displacement is 1 Find the angular _ frequency and the ampli- tude of the motion in terms of the given quantities. Hint: Since this is SHO, energy is conserved<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To solve this, we use the <strong>principle of conservation of energy<\/strong> for a particle in <strong>Simple Harmonic Motion (SHM)<\/strong>. The total mechanical energy remains constant and is the sum of kinetic and potential energy at any point.<\/p>\n\n\n\n<p>Let:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>v1v_1v1\u200b be the velocity when displacement is x1x_1x1\u200b<\/li>\n\n\n\n<li>v2v_2v2\u200b be the velocity when displacement is x2x_2x2\u200b<\/li>\n\n\n\n<li>\u03c9\\omega\u03c9 be the angular frequency<\/li>\n\n\n\n<li>AAA be the amplitude<\/li>\n\n\n\n<li>mmm be the mass of the particle (it will cancel out, so its value is not needed)<\/li>\n<\/ul>\n\n\n\n<p>The total energy in SHM is given by:E=12mv2+12m\u03c92&#215;2=12m\u03c92A2E = \\frac{1}{2}mv^2 + \\frac{1}{2}m\\omega^2 x^2 = \\frac{1}{2}m\\omega^2 A^2E=21\u200bmv2+21\u200bm\u03c92&#215;2=21\u200bm\u03c92A2<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Write energy expressions at two positions<\/h3>\n\n\n\n<p>At position x1x_1x1\u200b, the total energy is:12mv12+12m\u03c92&#215;12\\frac{1}{2}mv_1^2 + \\frac{1}{2}m\\omega^2 x_1^221\u200bmv12\u200b+21\u200bm\u03c92&#215;12\u200b<\/p>\n\n\n\n<p>At position x2x_2x2\u200b, the total energy is:12mv22+12m\u03c92&#215;22\\frac{1}{2}mv_2^2 + \\frac{1}{2}m\\omega^2 x_2^221\u200bmv22\u200b+21\u200bm\u03c92&#215;22\u200b<\/p>\n\n\n\n<p>Since total energy is conserved:12mv12+12m\u03c92&#215;12=12mv22+12m\u03c92&#215;22\\frac{1}{2}mv_1^2 + \\frac{1}{2}m\\omega^2 x_1^2 = \\frac{1}{2}mv_2^2 + \\frac{1}{2}m\\omega^2 x_2^221\u200bmv12\u200b+21\u200bm\u03c92&#215;12\u200b=21\u200bmv22\u200b+21\u200bm\u03c92&#215;22\u200b<\/p>\n\n\n\n<p>Divide through by 12m\\frac{1}{2}m21\u200bm:v12+\u03c92&#215;12=v22+\u03c92x22v_1^2 + \\omega^2 x_1^2 = v_2^2 + \\omega^2 x_2^2v12\u200b+\u03c92&#215;12\u200b=v22\u200b+\u03c92&#215;22\u200b<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Solve for \u03c92\\omega^2\u03c92<\/h3>\n\n\n\n<p>\u03c92(x12\u2212x22)=v22\u2212v12\\omega^2(x_1^2 &#8211; x_2^2) = v_2^2 &#8211; v_1^2\u03c92(x12\u200b\u2212x22\u200b)=v22\u200b\u2212v12\u200b\u03c92=v22\u2212v12x12\u2212x22\\omega^2 = \\frac{v_2^2 &#8211; v_1^2}{x_1^2 &#8211; x_2^2}\u03c92=x12\u200b\u2212x22\u200bv22\u200b\u2212v12\u200b\u200b<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Find Amplitude AAA<\/h3>\n\n\n\n<p>From total energy:12m\u03c92A2=12mv12+12m\u03c92&#215;12\\frac{1}{2}m\\omega^2 A^2 = \\frac{1}{2}mv_1^2 + \\frac{1}{2}m\\omega^2 x_1^221\u200bm\u03c92A2=21\u200bmv12\u200b+21\u200bm\u03c92&#215;12\u200b<\/p>\n\n\n\n<p>Divide by 12m\\frac{1}{2}m21\u200bm:\u03c92A2=v12+\u03c92&#215;12\\omega^2 A^2 = v_1^2 + \\omega^2 x_1^2\u03c92A2=v12\u200b+\u03c92&#215;12\u200bA2=v12\u03c92+x12A^2 = \\frac{v_1^2}{\\omega^2} + x_1^2A2=\u03c92v12\u200b\u200b+x12\u200b<\/p>\n\n\n\n<p>So,A=x12+v12\u03c92A = \\sqrt{x_1^2 + \\frac{v_1^2}{\\omega^2}}A=x12\u200b+\u03c92v12\u200b\u200b\u200b<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answers:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Angular frequency:<\/li>\n<\/ul>\n\n\n\n<p>\u03c9=v22\u2212v12x12\u2212x22\\omega = \\sqrt{ \\frac{v_2^2 &#8211; v_1^2}{x_1^2 &#8211; x_2^2} }\u03c9=x12\u200b\u2212x22\u200bv22\u200b\u2212v12\u200b\u200b\u200b<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Amplitude:<\/li>\n<\/ul>\n\n\n\n<p>A=x12+v12\u03c92A = \\sqrt{x_1^2 + \\frac{v_1^2}{\\omega^2}}A=x12\u200b+\u03c92v12\u200b\u200b\u200b<\/p>\n\n\n\n<p>This method relies purely on energy conservation and algebraic manipulation, which is valid for any SHM system where velocity and displacement are known at two points.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"852\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-735.jpeg\" alt=\"\" class=\"wp-image-36034\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-735.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-735-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-735-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>A particle undergoing simple harmonic motion has a velocity 1, when the displacement is X1 anda velocity %, when the displacement is 1 Find the angular _ frequency and the ampli- tude of the motion in terms of the given quantities. Hint: Since this is SHO, energy is conserved The Correct Answer and Explanation is: [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-36033","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36033","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=36033"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36033\/revisions"}],"predecessor-version":[{"id":36035,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36033\/revisions\/36035"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=36033"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=36033"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=36033"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}