{"id":87471,"date":"2025-06-01T06:44:50","date_gmt":"2025-06-01T06:44:50","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=87471"},"modified":"2025-06-01T06:44:50","modified_gmt":"2025-06-01T06:44:50","slug":"which-one-of-the-following-statements-is-correct-regarding-the-provide","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-statements-is-correct-regarding-the-provide\/","title":{"rendered":"Which one of the following statements is correct regarding the provide"},"content":{"rendered":"

Which one of the following statements is correct regarding the provided displacement versus time curve for a particle executing simple harmonic motion?<\/p>\n

[amp_mcq option1=”Phase of the oscillating particle is same at t=1s and t=3s” option2=”Phase of the oscillating particle is same at t=2s and t=8s” option3=”Phase of the oscillating particle is same at t=3s and t=17s” option4=”Phase of the oscillating particle is same at t=4s and t=10s” correct=”option2″]<\/p>\n

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This question was previously asked in<\/div>\n
UPSC NDA-1 – 2017<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nFor a particle executing simple harmonic motion (SHM), the phase of oscillation at time $t$ is given by $\\phi(t) = \\omega t + \\phi_0$, where $\\omega = 2\\pi\/T$ is the angular frequency and $T$ is the period. The phase is the same at two different times $t_1$ and $t_2$ if their phase difference is an integer multiple of $2\\pi$, i.e., $\\phi(t_2) – \\phi(t_1) = 2n\\pi$, which simplifies to $\\omega(t_2 – t_1) = 2n\\pi$, or $(t_2 – t_1) = nT$, where $n$ is an integer. This means the phase is the same at times separated by an integer multiple of the period T. In option B, the time difference is $8s – 2s = 6s$. If the period T of the motion is a divisor of 6s (e.g., T=6s, T=3s, T=2s, T=1s), then 6s is an integer multiple of T, and the phase would be the same at these times. Assuming the question is valid and option B is the intended correct answer, it implies that 6s is an integer multiple of the period T, while the time differences in options A (2s) and C (14s) are not integer multiples of T. (Note: Option D also has a time difference of 6s, which could imply it is also correct if the period divides 6s. Without the actual graph, definitively determining the period and confirming only B is correct is not possible. However, based on the provided answer choices and typical SHM properties, a period that makes 6s a multiple is implied).
\n<\/section>\n
\nThe phase of simple harmonic motion repeats itself every period T. Therefore, points in time separated by an integer multiple of T have the same phase.
\n<\/section>\n
\nThe displacement, velocity, and acceleration of a particle in SHM are all periodic functions with the same period T. The phase determines the instantaneous state of the oscillation (displacement, velocity, and direction of motion).
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

Which one of the following statements is correct regarding the provided displacement versus time curve for a particle executing simple harmonic motion? [amp_mcq option1=”Phase of the oscillating particle is same at t=1s and t=3s” option2=”Phase of the oscillating particle is same at t=2s and t=8s” option3=”Phase of the oscillating particle is same at t=3s and … <\/p>\n

Detailed SolutionWhich one of the following statements is correct regarding the provide<\/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":[1093],"tags":[1101,1129,1128],"class_list":["post-87471","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-1","tag-1101","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nWhich one of the following statements is correct regarding the provide<\/title>\n<meta name=\"description\" content=\"For a particle executing simple harmonic motion (SHM), the phase of oscillation at time $t$ is given by $phi(t) = omega t + phi_0$, where $omega = 2pi\/T$ is the angular frequency and $T$ is the period. The phase is the same at two different times $t_1$ and $t_2$ if their phase difference is an integer multiple of $2pi$, i.e., $phi(t_2) - phi(t_1) = 2npi$, which simplifies to $omega(t_2 - t_1) = 2npi$, or $(t_2 - t_1) = nT$, where $n$ is an integer. This means the phase is the same at times separated by an integer multiple of the period T. In option B, the time difference is $8s - 2s = 6s$. If the period T of the motion is a divisor of 6s (e.g., T=6s, T=3s, T=2s, T=1s), then 6s is an integer multiple of T, and the phase would be the same at these times. Assuming the question is valid and option B is the intended correct answer, it implies that 6s is an integer multiple of the period T, while the time differences in options A (2s) and C (14s) are not integer multiples of T. (Note: Option D also has a time difference of 6s, which could imply it is also correct if the period divides 6s. Without the actual graph, definitively determining the period and confirming only B is correct is not possible. However, based on the provided answer choices and typical SHM properties, a period that makes 6s a multiple is implied). The phase of simple harmonic motion repeats itself every period T. Therefore, points in time separated by an integer multiple of T have the same phase.\" \/>\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\/which-one-of-the-following-statements-is-correct-regarding-the-provide\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Which one of the following statements is correct regarding the provide\" \/>\n<meta property=\"og:description\" content=\"For a particle executing simple harmonic motion (SHM), the phase of oscillation at time $t$ is given by $phi(t) = omega t + phi_0$, where $omega = 2pi\/T$ is the angular frequency and $T$ is the period. The phase is the same at two different times $t_1$ and $t_2$ if their phase difference is an integer multiple of $2pi$, i.e., $phi(t_2) - phi(t_1) = 2npi$, which simplifies to $omega(t_2 - t_1) = 2npi$, or $(t_2 - t_1) = nT$, where $n$ is an integer. This means the phase is the same at times separated by an integer multiple of the period T. In option B, the time difference is $8s - 2s = 6s$. If the period T of the motion is a divisor of 6s (e.g., T=6s, T=3s, T=2s, T=1s), then 6s is an integer multiple of T, and the phase would be the same at these times. Assuming the question is valid and option B is the intended correct answer, it implies that 6s is an integer multiple of the period T, while the time differences in options A (2s) and C (14s) are not integer multiples of T. (Note: Option D also has a time difference of 6s, which could imply it is also correct if the period divides 6s. Without the actual graph, definitively determining the period and confirming only B is correct is not possible. However, based on the provided answer choices and typical SHM properties, a period that makes 6s a multiple is implied). The phase of simple harmonic motion repeats itself every period T. Therefore, points in time separated by an integer multiple of T have the same phase.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-statements-is-correct-regarding-the-provide\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T06:44:50+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=\"2 minutes\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"Which one of the following statements is correct regarding the provide","description":"For a particle executing simple harmonic motion (SHM), the phase of oscillation at time $t$ is given by $phi(t) = omega t + phi_0$, where $omega = 2pi\/T$ is the angular frequency and $T$ is the period. The phase is the same at two different times $t_1$ and $t_2$ if their phase difference is an integer multiple of $2pi$, i.e., $phi(t_2) - phi(t_1) = 2npi$, which simplifies to $omega(t_2 - t_1) = 2npi$, or $(t_2 - t_1) = nT$, where $n$ is an integer. This means the phase is the same at times separated by an integer multiple of the period T. In option B, the time difference is $8s - 2s = 6s$. If the period T of the motion is a divisor of 6s (e.g., T=6s, T=3s, T=2s, T=1s), then 6s is an integer multiple of T, and the phase would be the same at these times. Assuming the question is valid and option B is the intended correct answer, it implies that 6s is an integer multiple of the period T, while the time differences in options A (2s) and C (14s) are not integer multiples of T. (Note: Option D also has a time difference of 6s, which could imply it is also correct if the period divides 6s. Without the actual graph, definitively determining the period and confirming only B is correct is not possible. However, based on the provided answer choices and typical SHM properties, a period that makes 6s a multiple is implied). The phase of simple harmonic motion repeats itself every period T. Therefore, points in time separated by an integer multiple of T have the same phase.","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\/which-one-of-the-following-statements-is-correct-regarding-the-provide\/","og_locale":"en_US","og_type":"article","og_title":"Which one of the following statements is correct regarding the provide","og_description":"For a particle executing simple harmonic motion (SHM), the phase of oscillation at time $t$ is given by $phi(t) = omega t + phi_0$, where $omega = 2pi\/T$ is the angular frequency and $T$ is the period. The phase is the same at two different times $t_1$ and $t_2$ if their phase difference is an integer multiple of $2pi$, i.e., $phi(t_2) - phi(t_1) = 2npi$, which simplifies to $omega(t_2 - t_1) = 2npi$, or $(t_2 - t_1) = nT$, where $n$ is an integer. This means the phase is the same at times separated by an integer multiple of the period T. In option B, the time difference is $8s - 2s = 6s$. If the period T of the motion is a divisor of 6s (e.g., T=6s, T=3s, T=2s, T=1s), then 6s is an integer multiple of T, and the phase would be the same at these times. Assuming the question is valid and option B is the intended correct answer, it implies that 6s is an integer multiple of the period T, while the time differences in options A (2s) and C (14s) are not integer multiples of T. (Note: Option D also has a time difference of 6s, which could imply it is also correct if the period divides 6s. Without the actual graph, definitively determining the period and confirming only B is correct is not possible. However, based on the provided answer choices and typical SHM properties, a period that makes 6s a multiple is implied). The phase of simple harmonic motion repeats itself every period T. Therefore, points in time separated by an integer multiple of T have the same phase.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-statements-is-correct-regarding-the-provide\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T06:44:50+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"2 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-statements-is-correct-regarding-the-provide\/","url":"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-statements-is-correct-regarding-the-provide\/","name":"Which one of the following statements is correct regarding the provide","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T06:44:50+00:00","dateModified":"2025-06-01T06:44:50+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"For a particle executing simple harmonic motion (SHM), the phase of oscillation at time $t$ is given by $\\phi(t) = \\omega t + \\phi_0$, where $\\omega = 2\\pi\/T$ is the angular frequency and $T$ is the period. The phase is the same at two different times $t_1$ and $t_2$ if their phase difference is an integer multiple of $2\\pi$, i.e., $\\phi(t_2) - \\phi(t_1) = 2n\\pi$, which simplifies to $\\omega(t_2 - t_1) = 2n\\pi$, or $(t_2 - t_1) = nT$, where $n$ is an integer. This means the phase is the same at times separated by an integer multiple of the period T. In option B, the time difference is $8s - 2s = 6s$. If the period T of the motion is a divisor of 6s (e.g., T=6s, T=3s, T=2s, T=1s), then 6s is an integer multiple of T, and the phase would be the same at these times. Assuming the question is valid and option B is the intended correct answer, it implies that 6s is an integer multiple of the period T, while the time differences in options A (2s) and C (14s) are not integer multiples of T. (Note: Option D also has a time difference of 6s, which could imply it is also correct if the period divides 6s. Without the actual graph, definitively determining the period and confirming only B is correct is not possible. However, based on the provided answer choices and typical SHM properties, a period that makes 6s a multiple is implied). The phase of simple harmonic motion repeats itself every period T. Therefore, points in time separated by an integer multiple of T have the same phase.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-statements-is-correct-regarding-the-provide\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-statements-is-correct-regarding-the-provide\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-statements-is-correct-regarding-the-provide\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC NDA-1","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-nda-1\/"},{"@type":"ListItem","position":3,"name":"Which one of the following statements is correct regarding the provide"}]},{"@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\/87471","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=87471"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/87471\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=87471"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=87471"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=87471"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}