{"id":88289,"date":"2025-06-01T07:07:11","date_gmt":"2025-06-01T07:07:11","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=88289"},"modified":"2025-06-01T07:07:11","modified_gmt":"2025-06-01T07:07:11","slug":"a-particle-executes-linear-simple-harmonic-motion-with-amplitude-of-2","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-particle-executes-linear-simple-harmonic-motion-with-amplitude-of-2\/","title":{"rendered":"A particle executes linear simple harmonic motion with amplitude of 2"},"content":{"rendered":"

A particle executes linear simple harmonic motion with amplitude of 2 cm. When the particle is at 1 cm from the mean position, the magnitudes of the velocity and the acceleration are equal. Then its time period (in seconds) is:<\/p>\n

[amp_mcq option1=”2\u03c0” option2=”2\u03c0\/\u221a3″ option3=”\u221a3\/2\u03c0” option4=”2\u03c0\u221a3″ correct=”option2″]<\/p>\n

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This question was previously asked in<\/div>\n
UPSC NDA-2 – 2016<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
For a particle executing simple harmonic motion (SHM) with amplitude $A$ and angular frequency $\\omega$, the velocity $v$ at a displacement $x$ from the mean position is given by $v = \\pm \\omega \\sqrt{A^2 – x^2}$, and the acceleration $a$ is given by $a = -\\omega^2 x$. The magnitudes are $|v| = \\omega \\sqrt{A^2 – x^2}$ and $|a| = \\omega^2 |x|$. Given $A=2$ cm, and at $x=1$ cm, $|v| = |a|$.
\nSubstituting the values: $\\omega \\sqrt{2^2 – 1^2} = \\omega^2 |1|$
\n$\\omega \\sqrt{4 – 1} = \\omega^2$
\n$\\omega \\sqrt{3} = \\omega^2$
\nSince $\\omega$ for SHM is non-zero, we can divide by $\\omega$:
\n$\\sqrt{3} = \\omega$.
\nThe time period $T$ is related to angular frequency $\\omega$ by $T = \\frac{2\\pi}{\\omega}$.
\n$T = \\frac{2\\pi}{\\sqrt{3}}$.<\/section>\n
Formulas for velocity and acceleration in SHM are $|v| = \\omega \\sqrt{A^2 – x^2}$ and $|a| = \\omega^2 |x|$. The time period is $T = 2\\pi\/\\omega$.<\/section>\n
The velocity is maximum at the mean position ($x=0$) and zero at the extreme positions ($x=\\pm A$). The acceleration is maximum at the extreme positions and zero at the mean position. The relationship between velocity and displacement is elliptical in phase space, while the relationship between acceleration and displacement is linear.<\/section>\n","protected":false},"excerpt":{"rendered":"

A particle executes linear simple harmonic motion with amplitude of 2 cm. When the particle is at 1 cm from the mean position, the magnitudes of the velocity and the acceleration are equal. Then its time period (in seconds) is: [amp_mcq option1=”2\u03c0” option2=”2\u03c0\/\u221a3″ option3=”\u221a3\/2\u03c0” option4=”2\u03c0\u221a3″ correct=”option2″] This question was previously asked in UPSC NDA-2 – … <\/p>\n

Detailed SolutionA particle executes linear simple harmonic motion with amplitude of 2<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1094],"tags":[1098,1129,1128],"class_list":["post-88289","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-2","tag-1098","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nA particle executes linear simple harmonic motion with amplitude of 2<\/title>\n<meta name=\"description\" content=\"For a particle executing simple harmonic motion (SHM) with amplitude $A$ and angular frequency $omega$, the velocity $v$ at a displacement $x$ from the mean position is given by $v = pm omega sqrt{A^2 - x^2}$, and the acceleration $a$ is given by $a = -omega^2 x$. The magnitudes are $|v| = omega sqrt{A^2 - x^2}$ and $|a| = omega^2 |x|$. Given $A=2$ cm, and at $x=1$ cm, $|v| = |a|$. Substituting the values: $omega sqrt{2^2 - 1^2} = omega^2 |1|$ $omega sqrt{4 - 1} = omega^2$ $omega sqrt{3} = omega^2$ Since $omega$ for SHM is non-zero, we can divide by $omega$: $sqrt{3} = omega$. The time period $T$ is related to angular frequency $omega$ by $T = frac{2pi}{omega}$. $T = frac{2pi}{sqrt{3}}$. Formulas for velocity and acceleration in SHM are $|v| = omega sqrt{A^2 - x^2}$ and $|a| = omega^2 |x|$. The time period is $T = 2pi\/omega$.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/a-particle-executes-linear-simple-harmonic-motion-with-amplitude-of-2\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A particle executes linear simple harmonic motion with amplitude of 2\" \/>\n<meta property=\"og:description\" content=\"For a particle executing simple harmonic motion (SHM) with amplitude $A$ and angular frequency $omega$, the velocity $v$ at a displacement $x$ from the mean position is given by $v = pm omega sqrt{A^2 - x^2}$, and the acceleration $a$ is given by $a = -omega^2 x$. The magnitudes are $|v| = omega sqrt{A^2 - x^2}$ and $|a| = omega^2 |x|$. Given $A=2$ cm, and at $x=1$ cm, $|v| = |a|$. Substituting the values: $omega sqrt{2^2 - 1^2} = omega^2 |1|$ $omega sqrt{4 - 1} = omega^2$ $omega sqrt{3} = omega^2$ Since $omega$ for SHM is non-zero, we can divide by $omega$: $sqrt{3} = omega$. The time period $T$ is related to angular frequency $omega$ by $T = frac{2pi}{omega}$. $T = frac{2pi}{sqrt{3}}$. Formulas for velocity and acceleration in SHM are $|v| = omega sqrt{A^2 - x^2}$ and $|a| = omega^2 |x|$. The time period is $T = 2pi\/omega$.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-particle-executes-linear-simple-harmonic-motion-with-amplitude-of-2\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T07:07:11+00:00\" \/>\n<meta name=\"author\" content=\"rawan239\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rawan239\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"1 minute\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"A particle executes linear simple harmonic motion with amplitude of 2","description":"For a particle executing simple harmonic motion (SHM) with amplitude $A$ and angular frequency $omega$, the velocity $v$ at a displacement $x$ from the mean position is given by $v = pm omega sqrt{A^2 - x^2}$, and the acceleration $a$ is given by $a = -omega^2 x$. The magnitudes are $|v| = omega sqrt{A^2 - x^2}$ and $|a| = omega^2 |x|$. Given $A=2$ cm, and at $x=1$ cm, $|v| = |a|$. Substituting the values: $omega sqrt{2^2 - 1^2} = omega^2 |1|$ $omega sqrt{4 - 1} = omega^2$ $omega sqrt{3} = omega^2$ Since $omega$ for SHM is non-zero, we can divide by $omega$: $sqrt{3} = omega$. The time period $T$ is related to angular frequency $omega$ by $T = frac{2pi}{omega}$. $T = frac{2pi}{sqrt{3}}$. Formulas for velocity and acceleration in SHM are $|v| = omega sqrt{A^2 - x^2}$ and $|a| = omega^2 |x|$. The time period is $T = 2pi\/omega$.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/a-particle-executes-linear-simple-harmonic-motion-with-amplitude-of-2\/","og_locale":"en_US","og_type":"article","og_title":"A particle executes linear simple harmonic motion with amplitude of 2","og_description":"For a particle executing simple harmonic motion (SHM) with amplitude $A$ and angular frequency $omega$, the velocity $v$ at a displacement $x$ from the mean position is given by $v = pm omega sqrt{A^2 - x^2}$, and the acceleration $a$ is given by $a = -omega^2 x$. The magnitudes are $|v| = omega sqrt{A^2 - x^2}$ and $|a| = omega^2 |x|$. Given $A=2$ cm, and at $x=1$ cm, $|v| = |a|$. Substituting the values: $omega sqrt{2^2 - 1^2} = omega^2 |1|$ $omega sqrt{4 - 1} = omega^2$ $omega sqrt{3} = omega^2$ Since $omega$ for SHM is non-zero, we can divide by $omega$: $sqrt{3} = omega$. The time period $T$ is related to angular frequency $omega$ by $T = frac{2pi}{omega}$. $T = frac{2pi}{sqrt{3}}$. Formulas for velocity and acceleration in SHM are $|v| = omega sqrt{A^2 - x^2}$ and $|a| = omega^2 |x|$. The time period is $T = 2pi\/omega$.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-particle-executes-linear-simple-harmonic-motion-with-amplitude-of-2\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T07:07:11+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"1 minute"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-particle-executes-linear-simple-harmonic-motion-with-amplitude-of-2\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-particle-executes-linear-simple-harmonic-motion-with-amplitude-of-2\/","name":"A particle executes linear simple harmonic motion with amplitude of 2","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T07:07:11+00:00","dateModified":"2025-06-01T07:07:11+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"For a particle executing simple harmonic motion (SHM) with amplitude $A$ and angular frequency $\\omega$, the velocity $v$ at a displacement $x$ from the mean position is given by $v = \\pm \\omega \\sqrt{A^2 - x^2}$, and the acceleration $a$ is given by $a = -\\omega^2 x$. The magnitudes are $|v| = \\omega \\sqrt{A^2 - x^2}$ and $|a| = \\omega^2 |x|$. Given $A=2$ cm, and at $x=1$ cm, $|v| = |a|$. Substituting the values: $\\omega \\sqrt{2^2 - 1^2} = \\omega^2 |1|$ $\\omega \\sqrt{4 - 1} = \\omega^2$ $\\omega \\sqrt{3} = \\omega^2$ Since $\\omega$ for SHM is non-zero, we can divide by $\\omega$: $\\sqrt{3} = \\omega$. The time period $T$ is related to angular frequency $\\omega$ by $T = \\frac{2\\pi}{\\omega}$. $T = \\frac{2\\pi}{\\sqrt{3}}$. Formulas for velocity and acceleration in SHM are $|v| = \\omega \\sqrt{A^2 - x^2}$ and $|a| = \\omega^2 |x|$. The time period is $T = 2\\pi\/\\omega$.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/a-particle-executes-linear-simple-harmonic-motion-with-amplitude-of-2\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/a-particle-executes-linear-simple-harmonic-motion-with-amplitude-of-2\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-particle-executes-linear-simple-harmonic-motion-with-amplitude-of-2\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC NDA-2","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-nda-2\/"},{"@type":"ListItem","position":3,"name":"A particle executes linear simple harmonic motion with amplitude of 2"}]},{"@type":"WebSite","@id":"https:\/\/exam.pscnotes.com\/mcq\/#website","url":"https:\/\/exam.pscnotes.com\/mcq\/","name":"MCQ and Quiz for Exams","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/exam.pscnotes.com\/mcq\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209","name":"rawan239","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","caption":"rawan239"},"sameAs":["https:\/\/exam.pscnotes.com"],"url":"https:\/\/exam.pscnotes.com\/mcq\/author\/rawan239\/"}]}},"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/88289","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/comments?post=88289"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/88289\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=88289"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=88289"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=88289"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}