{"id":88469,"date":"2025-06-01T07:11:49","date_gmt":"2025-06-01T07:11:49","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=88469"},"modified":"2025-06-01T07:11:49","modified_gmt":"2025-06-01T07:11:49","slug":"the-time-period-of-oscillation-of-a-simple-pendulum-having-length-l-an","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/the-time-period-of-oscillation-of-a-simple-pendulum-having-length-l-an\/","title":{"rendered":"The time period of oscillation of a simple pendulum having length L an"},"content":{"rendered":"

The time period of oscillation of a simple pendulum having length L and mass of the bob m is given as T. If the length of the pendulum is increased to 4L and the mass of the bob is increased to 2m, then which one of the following is the new time period of oscillation?<\/p>\n

[amp_mcq option1=”T” option2=”2T” option3=”4T” option4=”T\/2″ correct=”option2″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-2 – 2018<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe new time period of oscillation is 2T.
\n<\/section>\n
\nThe time period (T) of a simple pendulum is given by the formula T = 2\u03c0\u221a(L\/g), where L is the length of the pendulum and g is the acceleration due to gravity. The mass of the bob (m) does not affect the time period of a simple pendulum.
\n<\/section>\n
\nLet the initial time period be T\u2081 = T, with length L\u2081 = L and mass m\u2081 = m. The formula is T\u2081 = 2\u03c0\u221a(L\/g).
\nWhen the length is increased to L\u2082 = 4L and the mass is increased to m\u2082 = 2m, the new time period T\u2082 is calculated using the formula, considering only the change in length:
\nT\u2082 = 2\u03c0\u221a(L\u2082\/g) = 2\u03c0\u221a((4L)\/g) = 2\u03c0 * \u221a4 * \u221a(L\/g) = 2\u03c0 * 2 * \u221a(L\/g) = 2 * (2\u03c0\u221a(L\/g)).
\nSince T\u2081 = 2\u03c0\u221a(L\/g), the new time period T\u2082 = 2 * T\u2081. Thus, the new time period is 2T.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

The time period of oscillation of a simple pendulum having length L and mass of the bob m is given as T. If the length of the pendulum is increased to 4L and the mass of the bob is increased to 2m, then which one of the following is the new time period of oscillation? … <\/p>\n

Detailed SolutionThe time period of oscillation of a simple pendulum having length L an<\/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":[1114,1129,1128],"class_list":["post-88469","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-2","tag-1114","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nThe time period of oscillation of a simple pendulum having length L an<\/title>\n<meta name=\"description\" content=\"The new time period of oscillation is 2T. The time period (T) of a simple pendulum is given by the formula T = 2\u03c0\u221a(L\/g), where L is the length of the pendulum and g is the acceleration due to gravity. The mass of the bob (m) does not affect the time period of a simple pendulum.\" \/>\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\/the-time-period-of-oscillation-of-a-simple-pendulum-having-length-l-an\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"The time period of oscillation of a simple pendulum having length L an\" \/>\n<meta property=\"og:description\" content=\"The new time period of oscillation is 2T. The time period (T) of a simple pendulum is given by the formula T = 2\u03c0\u221a(L\/g), where L is the length of the pendulum and g is the acceleration due to gravity. 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