{"id":2137,"date":"2022-12-29T10:00:00","date_gmt":"2022-12-29T04:30:00","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=2137"},"modified":"2022-12-30T09:45:04","modified_gmt":"2022-12-30T04:15:04","slug":"relation-between-wave-velocity-frequency-wavelength","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/relation-between-wave-velocity-frequency-wavelength\/","title":{"rendered":"What Is The Relation Between Wave Velocity, Frequency And Wavelength"},"content":{"rendered":"
\\( \\text{wave velocity}=\\frac{\\text{distance covered}}{\\text{Time}\\,\\,\\text{taken}} \\) Example 1:<\/strong> \u00a0 \u00a0If 50 waves are produced in 2 seconds, what is its frequency ? Example 2:<\/strong> \u00a0 \u00a0A source produce 50 crests and 50 troughs in 0.5 second. Find the frequency. Example 3:<\/strong> \u00a0 \u00a0Sound waves travel with a speed of 330 m\/s. What is the wavelength of sound waves whose frequency is 550 Hz ? Example 4:<\/strong> \u00a0 \u00a0The wave length of sound emitted by a source is 1.7 \u00d7 10-2<\/sup> m. Calculate frequency of the sound, if its velocity is 343.4 ms-1<\/sup>. Example 5:<\/strong> \u00a0 \u00a0A wave pulse on a string moves a distance of 8m in 0.05 s. Example 6:<\/strong> \u00a0 \u00a0A person has a hearing range of 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies ? Take the speed of sound in air as 340 m\/s. Example 7:<\/strong> \u00a0 \u00a0A longitudinal wave is produced on a toy string. The wave travels at a speed of 30 cm\/s and the frequency of the wave is 20 Hz. What is the minimum separation between the consecutive compressions of the string ? Example 8:<\/strong> \u00a0 \u00a0Wave of frequency 200 Hz produced in a string is represented in figure. Find out the following Example 9:<\/strong> \u00a0 \u00a0A stone is dropped into a well 44.1 m deep. The sound of splash is heard 3.13 seconds after the stone is dropped. Calculate the velocity of sound in air. Relation Between Wave Velocity, Frequency And Wavelength\u00a0For A Periodic Wave Wave Velocity = Frequency \u00d7 Wave Length Relation Between Wave Velocity, Frequency And Wavelength Example Problems With Solutions Example 1: \u00a0 \u00a0If 50 waves are produced in 2 seconds, what is its frequency ? Solution:\u00a0 \u00a0 Frequency, Example 2: \u00a0 \u00a0A source produce 50 crests […]<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[404],"tags":[585,882,881,895,584],"yoast_head":"\n
\n\\( =\\frac{\\text{wave}\\,\\text{length}}{\\text{Time}\\,\\text{taken}} \\)
\n\\( \\text{or v}=\\frac{\\lambda }{T}\\text{ }……\\text{ (1)} \\)
\n\\( \\text{since }\\!\\!~\\!\\!\\text{ }v=\\frac{\\text{1}}{\\text{T}}\\text{, equation }\\left( \\text{1} \\right)\\text{ can also be written as} \\)
\n\\( \\text{v}=v\\lambda \\text{ }……\\text{ (2)} \\)
\nWave Velocity = Frequency \u00d7 Wave Length<\/strong><\/p>\nRelation Between Wave Velocity, Frequency And Wavelength Example Problems With Solutions<\/strong><\/h2>\n
\nSolution:<\/strong>\u00a0 \u00a0 Frequency,
\n\\( v=\\frac{\\text{Number}\\,\\text{of}\\,\\text{wave}\\,\\text{produced}}{\\text{Time}\\,\\text{taken}} \\)
\n\\( =\\frac{50}{2}=25\\text{ Hz} \\)<\/p>\n
\nSolution:<\/strong>\u00a0 \u00a0 1 crest and 1 trough = 1 wave
\n\u2234 50 crests and 50 troughs = 50 waves
\n\\( \\text{Now, Frequency, }v=~\\frac{\\text{Number}\\,\\text{of}\\,\\text{wave}}{\\text{Time}} \\)
\n\\( =\\frac{50}{0.5}=100\\text{ Hz} \\)<\/p>\n
\nSolution:<\/strong>\u00a0 \u00a0 Given velocity, v = 330 m\/s,
\nFrequency, \\(\\upsilon\\)\u00a0= 550 Hz
\n\\(\\therefore \\text{ }wavelength,~~\\text{ }\\lambda =\\frac{\\text{v}}{\\upsilon }\\)
\n\\( =\\frac{330}{550}=0.6\\text{ }m \\)<\/p>\n
\nSolution:<\/strong>\u00a0 \u00a0 The relation ship between velocity, frequency and wave length of a wave is given by the formula v = \\(\\upsilon\\)\u00a0\u00d7 \u03bb
\nHere, velocity, v = 343.4 ms-1<\/sup>
\nfrequency \\(\\upsilon\\)\u00a0= ?
\nand wavelength, \u03bb= 1.7 \u00d7 10-2<\/sup> m
\nSo, putting these values in the above formula, we get :
\n343.4 = \\(\\upsilon\\)\u00a0\u00d7 1.7 \u00d7 10-2<\/sup>
\n\\( v=\\frac{343.4}{1.7\\times {{10}^{-2}}} \\)
\n\\( =\\frac{343.4\\,\\times {{10}^{2}}}{1.7} \\)
\n= 2.02 \u00d7 104<\/sup> Hz
\nThus, the frequency of sound is 2.02 \u00d7 104<\/sup> hertz.<\/p>\n
\n(i) Calculate the velocity of the pulse.
\n(ii) What would be the wavelength of the wave on the same string, if its frequency is 200 Hz ?
\nSolution:<\/strong>\u00a0 \u00a0 (i) Velocity of the wave,
\n\\( \\text{v}=\\frac{\\text{Distance covered}}{\\text{Time}\\,\\text{taken}}=\\frac{8m}{0.05s}=160\\text{ m\/s} \\)
\n(ii) Periodic wave has the same velocity as that of the wave pulse on the same string.
\n\\( \\therefore \\text{ Wavelength, }\\lambda =~\\frac{\\text{v}}{v}=~\\frac{160\\,m\/s}{200\\,Hz}~=0.8\\text{m} \\)
\nThus, the wavelength of the wave is 0.8 m.<\/p>\n
\nSolution:<\/strong>\u00a0 \u00a0 Given : \\({{v}_{1}}\\)\u00a0= 20 Hz, V = 340 m\/s
\n\\( \\therefore \\text{ }{{\\lambda }_{1}}=\\frac{\\text{v}}{{{v}_{1}}}=\\frac{340}{20}=17\\text{ m} \\)
\n\\({{v}_{2}}\\) = 20 kHz = 20,000 Hz, v = 340 m\/s
\n\\( \\therefore \\text{ }{{\\lambda }_{2}}=\\frac{\\text{v}}{{{v}_{2}}}=\\frac{340}{20,000}=1.7\\times {{10}^{-2}}m=1.7cm \\)
\n\u2234 The typical wavelengths are 17 m and 1.7 cm.<\/p>\n
\nSol. Given, Velocity, v = 30 cm\/s
\nFrequency, \\(\\upsilon\\)\u00a0= 20 Hz
\nMinimum separation between the two consecutive compressions is equal to one wavelength \u03bb and
\n\\(\\lambda =\\frac{\\text{v}}{v}=\\frac{30\\,cm\/s}{20\\,Hz}=1.5\\text{ cm}\\)<\/p>\n
\n
\n(i) amplitude
\n(ii) wavelength
\n(iii) wave velocity
\nSolution:<\/strong>\u00a0 \u00a0 (i) Amplitude = Maximum displacement = 10 cm
\n(ii) Wavelength \u03bb = Distance between two successive crests = 40 cm
\n(iii) Now, frequency, n = 2 Hz
\nWavelength, \u03bb= 40cm = 0.4 m
\n\u2234 Wave velocity, v = \\(\\upsilon\\)\u03bb
\n= 200 \u00d7 0.4 m\/s
\n= 80m\/s<\/p>\n
\nSolution:<\/strong>\u00a0 \u00a0 First we calculate the time taken by the stone to reach the water level by using the relation:
\ns = ut + \\(\\frac { 1 }{ 2 }\\) gt2<\/sup>
\nHere s = 44.1 m, u = 0, g = 9.8 m\/s2<\/sup>
\n\u2234 44.1 = 0 \u00d7 t + \\(\\frac { 1 }{ 2 }\\) \u00d7 9.8 \u00d7 t2<\/sup>
\n\\( {{t}^{2}}=\\frac{44.1\\times 2}{9.8}=9 \\)
\nor t = 3 s
\nTime taken by the sound to reach the top of the well
\nt2<\/sub>\u00a0= 3.13 \u2013 3 = 0.13 s
\nNow, speed of sound
\n\\( \\frac{\\text{Distance}}{\\text{Time}}=\\frac{44.1\\,m}{0.13\\,s}=339.2\\text{ m\/s} \\)<\/p>\n","protected":false},"excerpt":{"rendered":"