{"id":189374,"date":"2025-02-09T06:43:55","date_gmt":"2025-02-09T06:43:55","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=189374"},"modified":"2025-02-09T06:44:37","modified_gmt":"2025-02-09T06:44:37","slug":"a-stretched-string-has-a-mass-per-unit-length-of-5-94-g-cm-and-a-tension-of-19-8-n","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/09\/a-stretched-string-has-a-mass-per-unit-length-of-5-94-g-cm-and-a-tension-of-19-8-n\/","title":{"rendered":"A stretched string has a mass per unit length of 5.94 g\/cm and a tension of 19.8 N"},"content":{"rendered":"\n<p>A stretched string has a mass per unit length of 5.94 g\/cm and a tension of 19.8 N. A sinusoidal wave on this string has an amplitude of 0.142 mm and a frequency of 159 Hz and is traveling in the negative direction of an x axis. If the wave equation is of the form y(x,t) = ym sin(kx + wt), what are (a) ym, (b) k, and (c) w, and (d) the correct choice of sign in front of w?<br>(a) &#8211; Exact number no tolerance,\u00c3\u201a\u00c2 (b) and (c) &#8211; The tolerance is \u00c3\u201a\u00c2\u00b1 1 in the 3rd significant digit<\/p>\n\n\n\n<p>The second pic is the available units for the answers<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-202.png\" alt=\"\" class=\"wp-image-189375\"\/><\/figure>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>To determine the parameters of the sinusoidal wave on the stretched string, we will calculate the amplitude (ym), wave number (k), angular frequency (\u03c9), and the correct sign in front of \u03c9 in the wave equation.<\/p>\n\n\n\n<p><strong>Given Data:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass per unit length (\u03bc) = 5.94 g\/cm = 0.0594 kg\/m<\/li>\n\n\n\n<li>Tension (T) = 19.8 N<\/li>\n\n\n\n<li>Amplitude (A) = 0.142 mm = 0.000142 m<\/li>\n\n\n\n<li>Frequency (f) = 159 Hz<\/li>\n<\/ul>\n\n\n\n<p><strong>(a) Amplitude (ym):<\/strong><\/p>\n\n\n\n<p>The amplitude of the wave is given as 0.142 mm, which is 0.000142 m. Therefore,<\/p>\n\n\n\n<p><strong>ym = 0.000142 m<\/strong><\/p>\n\n\n\n<p><strong>(b) Wave Number (k):<\/strong><\/p>\n\n\n\n<p>The wave number k is related to the wavelength \u03bb by the equation k = 2\u03c0\/\u03bb. To find \u03bb, we use the relationship between wave speed (v), frequency (f), and wavelength:<\/p>\n\n\n\n<p>v = f \u00d7 \u03bb<\/p>\n\n\n\n<p>First, calculate the wave speed (v) using the formula:<\/p>\n\n\n\n<p>v = \u221a(T \/ \u03bc)<\/p>\n\n\n\n<p>Substituting the given values:<\/p>\n\n\n\n<p>v = \u221a(19.8 N \/ 0.0594 kg\/m) \u2248 18.26 m\/s<\/p>\n\n\n\n<p>Now, calculate the wavelength (\u03bb):<\/p>\n\n\n\n<p>\u03bb = v \/ f = 18.26 m\/s \/ 159 Hz \u2248 0.114 m<\/p>\n\n\n\n<p>Finally, calculate the wave number (k):<\/p>\n\n\n\n<p>k = 2\u03c0 \/ \u03bb \u2248 2\u03c0 \/ 0.114 m \u2248 55.1 rad\/m<\/p>\n\n\n\n<p><strong>(c) Angular Frequency (\u03c9):<\/strong><\/p>\n\n\n\n<p>The angular frequency \u03c9 is related to the frequency f by the equation \u03c9 = 2\u03c0f. Substituting the given frequency:<\/p>\n\n\n\n<p>\u03c9 = 2\u03c0 \u00d7 159 Hz \u2248 999.6 rad\/s<\/p>\n\n\n\n<p><strong>(d) Correct Sign in Front of \u03c9:<\/strong><\/p>\n\n\n\n<p>The wave is traveling in the negative direction of the x-axis, which corresponds to a wave function of the form y(x,t) = ym sin(kx + \u03c9t). Therefore, the correct sign in front of \u03c9 is positive.<\/p>\n\n\n\n<p><strong>Summary of Results:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Amplitude (ym):<\/strong> 0.000142 m<\/li>\n\n\n\n<li><strong>Wave Number (k):<\/strong> 55.1 rad\/m<\/li>\n\n\n\n<li><strong>Angular Frequency (\u03c9):<\/strong> 999.6 rad\/s<\/li>\n\n\n\n<li><strong>Correct Sign in Front of \u03c9:<\/strong> Positive<\/li>\n<\/ul>\n\n\n\n<p>These calculations are based on standard wave equations and the relationships between wave parameters.<\/p>\n\n\n\n<p>For a visual explanation of how wave speed depends on tension and linear mass density, you might find the following video helpful:<\/p>\n\n\n\n<p>\ue200video\ue202Wave Speed on a Stretched String\ue202turn0search6\ue201<\/p>\n","protected":false},"excerpt":{"rendered":"<p>A stretched string has a mass per unit length of 5.94 g\/cm and a tension of 19.8 N. A sinusoidal wave on this string has an amplitude of 0.142 mm and a frequency of 159 Hz and is traveling in the negative direction of an x axis. If the wave equation is of the form [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-189374","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189374","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=189374"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189374\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=189374"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=189374"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=189374"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}