{"id":189593,"date":"2025-02-10T05:34:37","date_gmt":"2025-02-10T05:34:37","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=189593"},"modified":"2025-02-10T05:34:40","modified_gmt":"2025-02-10T05:34:40","slug":"for-a-given-string-under-constant-tension","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/10\/for-a-given-string-under-constant-tension\/","title":{"rendered":"For a given string under constant tension"},"content":{"rendered":"\n

a) For a given string under constant tension, find an expression for the frequency of the nth harmonic, f, given the wave speed and the length of the string L.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

b) Derive the expression, v = 2l_{n}*f_{n} where I is the shortest distance between nodes for theth harmonic. Write your answer here and in Question 1.4 for reference during the lab.
c) Refer to Section 1.1. In Figure 2. 1. suppose a total mass M = 150 grams is hanging on the end of the string. Assume the linear mass density of the string is mu = 4.1g \/ m_{e} and the length of the string is L = 1.6m
Determine the expected frequency f of the second harmonic (n = 2) Write your answer here and in Question 1.1 for reference during the lab.<\/p>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

Part (a): Frequency of the nth Harmonic<\/h3>\n\n\n\n

For a string under constant tension, the frequency ( f_n ) of the ( n )-th harmonic can be determined by considering the wave speed and the length of the string. The general wave speed equation for a string under tension is:<\/p>\n\n\n\n

[
v = \\sqrt{\\frac{T}{\\mu}}
]<\/p>\n\n\n\n

Where:<\/p>\n\n\n\n

    \n
  • ( v ) is the wave speed,<\/li>\n\n\n\n
  • ( T ) is the tension in the string,<\/li>\n\n\n\n
  • ( \\mu ) is the linear mass density of the string (mass per unit length).<\/li>\n<\/ul>\n\n\n\n

    For a string fixed at both ends, the fundamental frequency (first harmonic, ( n = 1 )) corresponds to a standing wave with one antinode in the middle and nodes at the ends. The wavelength of the fundamental is twice the length of the string, i.e., ( \\lambda_1 = 2L ).<\/p>\n\n\n\n

    For the ( n )-th harmonic, the wavelength ( \\lambda_n ) is given by:<\/p>\n\n\n\n

    [
    \\lambda_n = \\frac{2L}{n}
    ]<\/p>\n\n\n\n

    The frequency ( f_n ) of the ( n )-th harmonic is related to the wave speed ( v ) and the wavelength ( \\lambda_n ) by:<\/p>\n\n\n\n

    [
    f_n = \\frac{v}{\\lambda_n}
    ]<\/p>\n\n\n\n

    Substituting for ( \\lambda_n ):<\/p>\n\n\n\n

    [
    f_n = \\frac{v}{\\frac{2L}{n}} = \\frac{n v}{2L}
    ]<\/p>\n\n\n\n

    So, the expression for the frequency of the ( n )-th harmonic is:<\/p>\n\n\n\n

    [
    f_n = \\frac{n}{2L} \\sqrt{\\frac{T}{\\mu}}
    ]<\/p>\n\n\n\n

    Part (b): Derivation of ( v = 2l_n f_n )<\/h3>\n\n\n\n

    For the ( n )-th harmonic, the distance between adjacent nodes (also called the node-to-node length) is ( l_n ). The total length of the string ( L ) is related to the number of nodes ( n ). For the ( n )-th harmonic, there are ( n + 1 ) nodes, and the distance between adjacent nodes is:<\/p>\n\n\n\n

    [
    l_n = \\frac{L}{n}
    ]<\/p>\n\n\n\n

    The frequency ( f_n ) is related to the wave speed and the wavelength. From the previous equation for ( f_n ), we have:<\/p>\n\n\n\n

    [
    f_n = \\frac{n v}{2L}
    ]<\/p>\n\n\n\n

    Rearranging this equation gives:<\/p>\n\n\n\n

    [
    v = 2 L f_n
    ]<\/p>\n\n\n\n

    Substituting for ( l_n ) in the equation ( L = n l_n ), we get:<\/p>\n\n\n\n

    [
    v = 2 l_n f_n
    ]<\/p>\n\n\n\n

    This completes the derivation.<\/p>\n\n\n\n

    Part (c): Determining the Frequency of the Second Harmonic<\/h3>\n\n\n\n

    Given:<\/p>\n\n\n\n

      \n
    • Total mass, ( M = 150 \\, \\text{grams} = 0.15 \\, \\text{kg} ),<\/li>\n\n\n\n
    • Linear mass density, ( \\mu = 4.1 \\, \\text{g\/m} = 4.1 \\times 10^{-3} \\, \\text{kg\/m} ),<\/li>\n\n\n\n
    • Length of the string, ( L = 1.6 \\, \\text{m} ),<\/li>\n\n\n\n
    • For the second harmonic, ( n = 2 ).<\/li>\n<\/ul>\n\n\n\n

      First, we need to determine the tension in the string. The tension ( T ) is caused by the hanging mass, so:<\/p>\n\n\n\n

      [
      T = M g
      ]<\/p>\n\n\n\n

      Where ( g ) is the acceleration due to gravity (( 9.8 \\, \\text{m\/s}^2 )):<\/p>\n\n\n\n

      [
      T = 0.15 \\times 9.8 = 1.47 \\, \\text{N}
      ]<\/p>\n\n\n\n

      Now, we can calculate the wave speed using:<\/p>\n\n\n\n

      [
      v = \\sqrt{\\frac{T}{\\mu}} = \\sqrt{\\frac{1.47}{4.1 \\times 10^{-3}}} = \\sqrt{358.54} = 18.94 \\, \\text{m\/s}
      ]<\/p>\n\n\n\n

      Now, using the expression for the frequency of the ( n )-th harmonic:<\/p>\n\n\n\n

      [
      f_n = \\frac{n}{2L} \\sqrt{\\frac{T}{\\mu}} = \\frac{2}{2 \\times 1.6} \\times 18.94 = \\frac{18.94}{1.6} = 11.84 \\, \\text{Hz}
      ]<\/p>\n\n\n\n

      Thus, the frequency of the second harmonic is approximately ( 11.84 \\, \\text{Hz} ).<\/p>\n\n\n\n

      Summary:<\/h3>\n\n\n\n
        \n
      • The frequency of the ( n )-th harmonic is given by: [
        f_n = \\frac{n}{2L} \\sqrt{\\frac{T}{\\mu}}
        ]<\/li>\n\n\n\n
      • The expression ( v = 2l_n f_n ) follows from the relationship between the wavelength and the length between nodes in the ( n )-th harmonic.<\/li>\n\n\n\n
      • For the second harmonic, the calculated frequency is approximately ( 11.84 \\, \\text{Hz} ).<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"

        a) For a given string under constant tension, find an expression for the frequency of the nth harmonic, f, given the wave speed and the length of the string L. b) Derive the expression, v = 2l_{n}*f_{n} where I is the shortest distance between nodes for theth harmonic. Write your answer here and in Question […]<\/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-189593","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189593","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=189593"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189593\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=189593"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=189593"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=189593"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}