{"id":38839,"date":"2025-06-26T13:30:28","date_gmt":"2025-06-26T13:30:28","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=38839"},"modified":"2025-06-26T13:30:30","modified_gmt":"2025-06-26T13:30:30","slug":"a-stretched-string-has-a-mass-per-unit-length-of-5-0-g-cm-and-a-tension-of-15-n","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/a-stretched-string-has-a-mass-per-unit-length-of-5-0-g-cm-and-a-tension-of-15-n\/","title":{"rendered":"A stretched string has a mass per unit length of 5.0 g\/cm and a tension of 15 N."},"content":{"rendered":"\n<p>A stretched string has a mass per unit length of 5.0 g\/cm and a tension of 15 N. A wave on this string has an amplitude of 0.13 mm and a frequency of 200 Hz and is traveling in the positive x direction. (a) What is the wave speed? (b) Write an equation for this wave (c) What is the wavelength of the traveling wave? (d) What is the maximum speed of a particle on the string?<\/p>\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>We are given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass per unit length: \u03bc = 5.0 g\/cm = 0.005 kg\/m<\/li>\n\n\n\n<li>Tension: T = 15 N<\/li>\n\n\n\n<li>Frequency: f = 200 Hz<\/li>\n\n\n\n<li>Amplitude: A = 0.13 mm = 1.3 \u00d7 10\u207b\u2074 m<\/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\"><strong>(a) What is the wave speed?<\/strong><\/h3>\n\n\n\n<p>The wave speed vvv on a string is given by:v=T\u03bcv = \\sqrt{\\frac{T}{\\mu}}v=\u03bcT\u200b\u200b<\/p>\n\n\n\n<p>Substitute the values:v=150.005=3000\u224854.77&nbsp;m\/sv = \\sqrt{\\frac{15}{0.005}} = \\sqrt{3000} \u2248 54.77\\ \\text{m\/s}v=0.00515\u200b\u200b=3000\u200b\u224854.77&nbsp;m\/s<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>(b) Write an equation for this wave<\/strong><\/h3>\n\n\n\n<p>A traveling wave moving in the positive x-direction can be written as:y(x,t)=Asin\u2061(kx\u2212\u03c9t)y(x, t) = A \\sin(kx &#8211; \\omega t)y(x,t)=Asin(kx\u2212\u03c9t)<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>A=1.3\u00d710\u22124A = 1.3 \\times 10^{-4}A=1.3\u00d710\u22124 m<\/li>\n\n\n\n<li>f=200f = 200f=200 Hz<\/li>\n\n\n\n<li>\u03c9=2\u03c0f=400\u03c0\u00a0rad\/s\\omega = 2\\pi f = 400\\pi \\ \\text{rad\/s}\u03c9=2\u03c0f=400\u03c0\u00a0rad\/s<\/li>\n\n\n\n<li>v=54.77\u00a0m\/sv = 54.77\\ \\text{m\/s}v=54.77\u00a0m\/s<\/li>\n\n\n\n<li>\u03bb=vf=54.77200\u22480.274\u00a0m\\lambda = \\frac{v}{f} = \\frac{54.77}{200} \u2248 0.274\\ \\text{m}\u03bb=fv\u200b=20054.77\u200b\u22480.274\u00a0m<\/li>\n\n\n\n<li>k=2\u03c0\u03bb=2\u03c00.274\u224822.94\u00a0rad\/mk = \\frac{2\\pi}{\\lambda} = \\frac{2\\pi}{0.274} \u2248 22.94\\ \\text{rad\/m}k=\u03bb2\u03c0\u200b=0.2742\u03c0\u200b\u224822.94\u00a0rad\/m<\/li>\n<\/ul>\n\n\n\n<p>So, the wave equation is:y(x,t)=(1.3\u00d710\u22124)sin\u2061(22.94x\u2212400\u03c0t)y(x, t) = (1.3 \\times 10^{-4}) \\sin(22.94x &#8211; 400\\pi t)y(x,t)=(1.3\u00d710\u22124)sin(22.94x\u2212400\u03c0t)<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>(c) What is the wavelength?<\/strong><\/h3>\n\n\n\n<p>\u03bb=vf=54.77200\u22480.274&nbsp;m\\lambda = \\frac{v}{f} = \\frac{54.77}{200} \u2248 0.274\\ \\text{m}\u03bb=fv\u200b=20054.77\u200b\u22480.274&nbsp;m<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>(d) What is the maximum speed of a particle on the string?<\/strong><\/h3>\n\n\n\n<p>The maximum particle speed is:vmax=\u03c9A=(400\u03c0)(1.3\u00d710\u22124)\u22480.163&nbsp;m\/sv_{\\text{max}} = \\omega A = (400\\pi)(1.3 \\times 10^{-4}) \u2248 0.163\\ \\text{m\/s}vmax\u200b=\u03c9A=(400\u03c0)(1.3\u00d710\u22124)\u22480.163&nbsp;m\/s<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation<\/strong><\/h3>\n\n\n\n<p>This problem explores the motion of a transverse wave on a stretched string. First, we calculate the <strong>wave speed<\/strong> using the formula v=T\/\u03bcv = \\sqrt{T\/\\mu}v=T\/\u03bc\u200b. The tension in the string provides the restoring force for wave motion, while the mass per unit length resists acceleration. A higher tension results in faster waves, while a higher mass per unit length slows them down. With a tension of 15 newtons and a linear mass density of 0.005 kilograms per meter, the wave speed is found to be approximately 54.77 meters per second.<\/p>\n\n\n\n<p>Next, we describe the wave mathematically. A sinusoidal traveling wave in the positive x-direction is represented as y(x,t)=Asin\u2061(kx\u2212\u03c9t)y(x, t) = A \\sin(kx &#8211; \\omega t)y(x,t)=Asin(kx\u2212\u03c9t). The amplitude AAA is how far particles on the string move up and down from equilibrium, here converted to meters. The angular frequency \u03c9\\omega\u03c9 is related to the frequency by \u03c9=2\u03c0f\\omega = 2\\pi f\u03c9=2\u03c0f, and the wave number kkk is calculated from the wavelength as k=2\u03c0\/\u03bbk = 2\\pi\/\\lambdak=2\u03c0\/\u03bb.<\/p>\n\n\n\n<p>To find the <strong>wavelength<\/strong>, we divide the wave speed by the frequency. This gives the distance between two adjacent crests or troughs. Lastly, the <strong>maximum speed<\/strong> of a particle on the string is given by vmax=\u03c9Av_{\\text{max}} = \\omega Avmax\u200b=\u03c9A, which indicates how fast a point on the string is moving up and down at its peak. This value is different from the wave speed, which is the speed of energy transfer along the string.<\/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-banner8-1069.jpeg\" alt=\"\" class=\"wp-image-38840\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1069.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1069-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1069-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>A stretched string has a mass per unit length of 5.0 g\/cm and a tension of 15 N. A wave on this string has an amplitude of 0.13 mm and a frequency of 200 Hz and is traveling in the positive x direction. (a) What is the wave speed? (b) Write an equation for this [&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-38839","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/38839","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=38839"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/38839\/revisions"}],"predecessor-version":[{"id":38841,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/38839\/revisions\/38841"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=38839"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=38839"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=38839"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}