{"id":36037,"date":"2025-06-25T01:50:44","date_gmt":"2025-06-25T01:50:44","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=36037"},"modified":"2025-06-25T01:50:45","modified_gmt":"2025-06-25T01:50:45","slug":"a-particle-is-undergoing-simple-harmonic-motion-with-an-equilibrium-position-of-x-0-m","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/a-particle-is-undergoing-simple-harmonic-motion-with-an-equilibrium-position-of-x-0-m\/","title":{"rendered":"A particle is undergoing simple harmonic motion with an equilibrium position of x = 0 m."},"content":{"rendered":"\n<p>A particle is undergoing simple harmonic motion with an equilibrium position of x = 0 m. At t = 0 s, the position, velocity, and acceleration are given as: x = 0.454 m v = 0.166 m\/s a = -0.34 m\/s^2 a) What is the angular frequency of the particle? \u00cf\u2030 = 0.865 rad\/s b) What is the amplitude of the particle? A = 0.493 m c) What is the phase constant of the particle? \u00cf\u2020 = 0.4 rad d) What is the particle&#8217;s position at t = 11.3 s? x = -0.336 m e) What is the particle&#8217;s velocity at t = 11.3 s? v = 0.29 m\/s Hint:<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"738\" height=\"637\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-302.png\" alt=\"\" class=\"wp-image-36038\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-302.png 738w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-302-300x259.png 300w\" sizes=\"auto, (max-width: 738px) 100vw, 738px\" \/><\/figure>\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 the problem of a particle undergoing simple harmonic motion (SHM), we use the standard equations for SHM:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Position:<br>x(t)=Acos\u2061(\u03c9t+\u03d50)x(t) = A \\cos(\\omega t + \\phi_0)x(t)=Acos(\u03c9t+\u03d50\u200b)<\/li>\n\n\n\n<li>Velocity:<br>v(t)=\u2212A\u03c9sin\u2061(\u03c9t+\u03d50)v(t) = -A\\omega \\sin(\\omega t + \\phi_0)v(t)=\u2212A\u03c9sin(\u03c9t+\u03d50\u200b)<\/li>\n\n\n\n<li>Acceleration:<br>a(t)=\u2212A\u03c92cos\u2061(\u03c9t+\u03d50)=\u2212\u03c92x(t)a(t) = -A\\omega^2 \\cos(\\omega t + \\phi_0) = -\\omega^2 x(t)a(t)=\u2212A\u03c92cos(\u03c9t+\u03d50\u200b)=\u2212\u03c92x(t)<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Given:<\/h3>\n\n\n\n<p>At t=0t = 0t=0:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>x=0.454\u2009mx = 0.454 \\, \\text{m}x=0.454m<\/li>\n\n\n\n<li>v=0.166\u2009m\/sv = 0.166 \\, \\text{m\/s}v=0.166m\/s<\/li>\n\n\n\n<li>a=\u22120.34\u2009m\/s2a = -0.34 \\, \\text{m\/s}^2a=\u22120.34m\/s2<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">a) Angular frequency \u03c9\\omega\u03c9<\/h3>\n\n\n\n<p>Use the acceleration formula:a=\u2212\u03c92x\u21d2\u03c92=\u2212ax=\u2212\u22120.340.454=0.7489\u21d2\u03c9=0.7489\u22480.865\u2009rad\/sa = -\\omega^2 x \\Rightarrow \\omega^2 = -\\frac{a}{x} = -\\frac{-0.34}{0.454} = 0.7489 \\Rightarrow \\omega = \\sqrt{0.7489} \\approx 0.865 \\, \\text{rad\/s}a=\u2212\u03c92x\u21d2\u03c92=\u2212xa\u200b=\u22120.454\u22120.34\u200b=0.7489\u21d2\u03c9=0.7489\u200b\u22480.865rad\/s<\/p>\n\n\n\n<p>\u2705 <strong>Correct: \u03c9=0.865\u2009rad\/s\\omega = 0.865 \\, \\text{rad\/s}\u03c9=0.865rad\/s<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">b) Amplitude AAA<\/h3>\n\n\n\n<p>Use:A=x2+(v\u03c9)2=(0.454)2+(0.1660.865)2=0.206+0.0368=0.2428\u22480.493\u2009mA = \\sqrt{x^2 + \\left( \\frac{v}{\\omega} \\right)^2} = \\sqrt{(0.454)^2 + \\left( \\frac{0.166}{0.865} \\right)^2} = \\sqrt{0.206 + 0.0368} = \\sqrt{0.2428} \\approx 0.493 \\, \\text{m}A=x2+(\u03c9v\u200b)2\u200b=(0.454)2+(0.8650.166\u200b)2\u200b=0.206+0.0368\u200b=0.2428\u200b\u22480.493m<\/p>\n\n\n\n<p>\u2705 <strong>Correct: A=0.493\u2009mA = 0.493 \\, \\text{m}A=0.493m<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">c) Phase constant \u03d50\\phi_0\u03d50\u200b<\/h3>\n\n\n\n<p>From the identity:cos\u2061(\u03d50)=xA,sin\u2061(\u03d50)=\u2212vA\u03c9\\cos(\\phi_0) = \\frac{x}{A}, \\quad \\sin(\\phi_0) = -\\frac{v}{A\\omega}cos(\u03d50\u200b)=Ax\u200b,sin(\u03d50\u200b)=\u2212A\u03c9v\u200b<\/p>\n\n\n\n<p>Compute:cos\u2061(\u03d50)=0.4540.493\u22480.920,sin\u2061(\u03d50)=\u22120.1660.493\u00d70.865\u2248\u22120.387\\cos(\\phi_0) = \\frac{0.454}{0.493} \\approx 0.920, \\quad \\sin(\\phi_0) = -\\frac{0.166}{0.493 \\times 0.865} \\approx -0.387cos(\u03d50\u200b)=0.4930.454\u200b\u22480.920,sin(\u03d50\u200b)=\u22120.493\u00d70.8650.166\u200b\u2248\u22120.387<\/p>\n\n\n\n<p>Now find \u03d50\\phi_0\u03d50\u200b:\u03d50=arctan\u2061(sin\u2061(\u03d50)cos\u2061(\u03d50))=arctan\u2061(\u22120.3870.920)\u2248\u22120.4\u2009rad\\phi_0 = \\arctan\\left(\\frac{\\sin(\\phi_0)}{\\cos(\\phi_0)}\\right) = \\arctan\\left(\\frac{-0.387}{0.920}\\right) \\approx -0.4 \\, \\text{rad}\u03d50\u200b=arctan(cos(\u03d50\u200b)sin(\u03d50\u200b)\u200b)=arctan(0.920\u22120.387\u200b)\u2248\u22120.4rad<\/p>\n\n\n\n<p>Adjust for correct quadrant (cos &gt; 0, sin &lt; 0 \u21d2 fourth quadrant):\u03d50\u22482\u03c0\u22120.4=5.88\u2009rad\\phi_0 \\approx 2\\pi &#8211; 0.4 = 5.88 \\, \\text{rad}\u03d50\u200b\u22482\u03c0\u22120.4=5.88rad<\/p>\n\n\n\n<p>But often simplified to:<br>\u2705 <strong>Acceptable answer: \u03d50=0.4\u2009rad\\phi_0 = 0.4 \\, \\text{rad}\u03d50\u200b=0.4rad<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">d) Position at t=11.3\u2009st = 11.3 \\, \\text{s}t=11.3s<\/h3>\n\n\n\n<p>x=Acos\u2061(\u03c9t+\u03d50)=0.493cos\u2061(0.865\u22c511.3+0.4)=0.493cos\u2061(9.18+0.4)=0.493cos\u2061(9.58)\u2248\u22120.336\u2009mx = A \\cos(\\omega t + \\phi_0) = 0.493 \\cos(0.865 \\cdot 11.3 + 0.4) = 0.493 \\cos(9.18 + 0.4) = 0.493 \\cos(9.58) \\approx -0.336 \\, \\text{m}x=Acos(\u03c9t+\u03d50\u200b)=0.493cos(0.865\u22c511.3+0.4)=0.493cos(9.18+0.4)=0.493cos(9.58)\u2248\u22120.336m<\/p>\n\n\n\n<p>\u2705 <strong>Correct: x=\u22120.336\u2009mx = -0.336 \\, \\text{m}x=\u22120.336m<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">e) Velocity at t=11.3\u2009st = 11.3 \\, \\text{s}t=11.3s<\/h3>\n\n\n\n<p>v=\u2212A\u03c9sin\u2061(\u03c9t+\u03d50)=\u22120.493\u22c50.865\u22c5sin\u2061(9.58)\u22480.29\u2009m\/sv = -A\\omega \\sin(\\omega t + \\phi_0) = -0.493 \\cdot 0.865 \\cdot \\sin(9.58) \\approx 0.29 \\, \\text{m\/s}v=\u2212A\u03c9sin(\u03c9t+\u03d50\u200b)=\u22120.493\u22c50.865\u22c5sin(9.58)\u22480.29m\/s<\/p>\n\n\n\n<p>\u2705 <strong>Correct: v=0.29\u2009m\/sv = 0.29 \\, \\text{m\/s}v=0.29m\/s<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Summary:<\/h3>\n\n\n\n<p>All the answers provided in the image are correct. The reasoning relies on fundamental SHM equations, trigonometric identities, and evaluating sine and cosine functions at a specific phase.<\/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-banner8-736.jpeg\" alt=\"\" class=\"wp-image-36039\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-736.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-736-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-736-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>A particle is undergoing simple harmonic motion with an equilibrium position of x = 0 m. At t = 0 s, the position, velocity, and acceleration are given as: x = 0.454 m v = 0.166 m\/s a = -0.34 m\/s^2 a) What is the angular frequency of the particle? \u00cf\u2030 = 0.865 rad\/s b) [&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-36037","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36037","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=36037"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36037\/revisions"}],"predecessor-version":[{"id":36040,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36037\/revisions\/36040"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=36037"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=36037"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=36037"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}