","comment":{"@type":"Comment","text":"Determine the wavelength of the vibrations by using the number of nodes and antinodes that generally comprise the length of a wave and then arrive at the appropriate frequency. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$4 Hz$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$0.04 Hz$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$0.5 Hz$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$5 Hz$$","position":1,"answerExplanation":{"@type":"Comment","text":" We know that, if $$\\lambda$$ is the wavelength of the vibrational pattern: $$ \\\\ 1 node = \\dfrac{1}{2}\\lambda = 1 antinode\\\\ $$ From the figure, we see that the vibrations of the $$ 75 cm$$ long antenna, $$ \\\\ $$ As we go from the fixed end to the free end progresses through one node and one half on an antinode, i.e., $$ \\\\ \\dfrac{1}{2}\\lambda + \\dfrac{1}{4}\\lambda = 75\\\\\\Rightarrow \\dfrac{(2+1)\\lambda}{4} = 75\\\\ \\Rightarrow \\lambda = 100 cm = 1 m\\\\ $$ Therefore, the frequency of these vibrations, given that $$ \\\\ v = 5 ms^{-1} \\text{is found to be:}\\\\ f = \\dfrac{v}{\\lambda}\\\\ \\Rightarrow f = \\dfrac{5}{1}\\\\ \\therefore f = 5 Hz$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Wave properties Quiz 1","text":"You decide to spend an exquisite evening at the Sydney Opera House on your trip to Australia and watch the Phantom of The Opera. The theme song of this theatrical play is known for its initial notes played on the pipe organ that embosses the overbearing tone of the scene onto the audience. If two open pipe organs when played together produce this note at the rate of three beats per second, find their frequency if one of the pipes is $$30 cm$$ long and the other is $$30.3 cm$$ long respectively. ","comment":{"@type":"Comment","text":"Begin by determining the speed of the wave propagated through both pipes. Recall that the fundamental frequency of an open organ pipe will be two times the frequency of a closed pipe. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$50 Hz$$ and $$100 Hz$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$303.2 Hz$$ and $$273.2 Hz$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$15.2 Hz$$ and $$15 Hz$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$151.5 Hz$$ and $$150 Hz$$","position":2,"answerExplanation":{"@type":"Comment","text":"Let the fundamental frequency of the two pipes be $$f$$ and $$f^{\\prime}$$ respectively. $$ \\\\ $$ We are given that $$ l = 30 cm$$ and $$l^{\\prime} = 30.3 cm$$. $$ \\\\ $$ The sound travels through both pipes with the same speed, say $$v$$. $$ \\\\ $$ So we have: $$ \\\\ f = \\dfrac{v}{2l}\\\\ \\Rightarrow f = \\dfrac{v}{2 \\times 30}\\\\ \\Rightarrow f = \\dfrac{v}{60}\\\\ $$ Similarly, $$ \\\\ f^{\\prime} = \\dfrac{v}{2l^{\\prime}}\\\\ \\Rightarrow f^{\\prime} = \\dfrac{v}{2\\times 30.3}\\\\ \\Rightarrow f^{\\prime} = \\dfrac{v}{60.6}\\\\ $$ Given that the two organs produce three beats per second when played together: $$ \\\\ f-f^{\\prime} = 3\\\\ \\Rightarrow \\dfrac{v}{60} - \\dfrac{v}{60.6} = 3\\\\ \\Rightarrow \\left(\\dfrac{60.6-60}{60 \\times 60.6}\\right)v = 3\\\\\\Rightarrow v = \\dfrac{3 \\times 60 \\times 60.6}{0.6}\\\\ $$ Plugging this value of $$v$$ into our frequency expressions we get: $$ \\\\ f = \\dfrac{3 \\times 60 \\times 60.6}{0.6 \\times 2 \\times 60} = 151.5 Hz\\\\ \\therefore f^{\\prime} = \\dfrac{3 \\times 60 \\times 60.6}{0.6 \\times 2 \\times 60.6} = 150 Hz$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Wave properties Quiz 1","text":"Suppose that you and your friend are kayaking in the shallow backwaters of Kerala and your kayaks are spaced at $$2 m$$ from each other when a motorboat zooms past at a distance, creating ripples that are propagating towards your kayaks. Your kayaks begin to bob up and down about $$ 5$$ times in a span of $$10 s$$ as the ripples pass. When your kayak was at a high point, your friend’s kayak was at a low point and vice versa with no crests between the two of you. Assuming that the ripples travelled in a direction parallel to the imaginary line connecting the two kayaks, determine the wavelength, frequency and speed with which the ripples travelled. ","comment":{"@type":"Comment","text":"Recall that a consecutive crest and trough together comprise a definitive length of any wave, and use this to determine the wavelength of the ripple. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$2 m$$, $$0.5 Hz$$, $$1 ms^{-1}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$2 m$$, $$1 Hz$$, $$2 ms^{-1}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$4 m$$, $$2 Hz$$, $$8 ms^{-1}$$","position":2}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$4 m$$, $$0.5 Hz$$, $$2 ms^{-1}$$","position":3,"answerExplanation":{"@type":"Comment","text":"We are given the distance between the two kayaks is $$2 m$$. $$ \\\\ $$ We know that, if $$\\lambda$$ is the wavelength of a wave, the distance between a consecutive crest and trough is one half the wavelength i.e.,when you are at a high point and your friend is at a low point: $$ \\\\ \\dfrac{1}{2}\\lambda = 2\\\\ \\Rightarrow \\lambda = 4 m\\\\ $$ Now, we are given that the kayak bobs up and down $$5 \\text{times in} 10 s$$ therefore, the frequency of the ripples will be: $$ \\\\ f = \\dfrac{5}{10} = 0.5 Hz\\\\ $$ Consequently, the speed with which the ripples travelled can be given as: $$ \\\\ v = f\\lambda\\\\ \\Rightarrow v = 0.5 \\times 4\\\\ \\therefore v = 2 ms^{-1}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Wave properties Quiz 1","text":"For the propagation of waves through a string, the tension of the string is essential for producing a disturbance in the string and the mass of the string is what drives the disturbance by lending it suitable kinetic energy. Suppose you have a stretched string of linear mass density $$ \\mu$$ possessing a tension T along its length, determine the speed of transverse waves that can be set up in this stretched string. Should you require the need for a constant of proportionality in your evaluation, set it to 1 as we do experimentally. ","comment":{"@type":"Comment","text":"Dimensionally analyse the quantities in such a way that whatever combination of proportionality you get should be homogenous with the dimensions of velocity. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\sqrt{\\dfrac{\\mu}{T}}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{T}{\\mu}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{\\mu}{T}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\sqrt{\\dfrac{T}{\\mu}}$$","position":2,"answerExplanation":{"@type":"Comment","text":"LEt $$a$$ and $$b$$ be any real numbers such that: $$ \\\\ v \\propto T^{a}\\mu^{b}\\\\ \\Rightarrow v = kT^a \\mu^b \\\\ $$ Setting $$k=1: \\\\ \\Rightarrow v = T^{a} \\mu^b\\\\ \\Rightarrow [v] = \\left[T^a \\mu^b\\right]\\\\ $$ Expressing the quantities in their dimensional form we get: $$($$since tension has dimensions of force and linear mass density is just mass per unit length$$)$$ $$ \\\\ \\left[LT^{-1}\\right] = \\left[MLT^{-2}\\right] ^a \\left[ML^{-1}\\right]^b\\\\ \\Rightarrow \\left[M^0 LT^{-1}\\right] = \\left[M^aL^aT^{-2a}\\right] \\left[M^bL^{-b}T^0\\right]\\\\ \\Rightarrow \\left[M^0 LT^{-1}\\right] = \\left[M^{a+b}L^{a-b}T^{-2a}\\right]\\\\ $$ Applying the principle of homogeneity of dimensions on both sides of the equation, we get: $$ \\\\ a+b = 0,a-b=1,-2a=1, $$ from which we get: $$ \\\\ a = \\dfrac{1}{2},b = -\\dfrac{1}{2}\\\\ $$ Plugging this back into our velocity equation we get: $$ \\\\ v = T ^{\\frac{1}{2}}\\mu ^{-\\frac{1}{2}}\\\\ \\therefore v = \\sqrt{\\dfrac{T}{\\mu}}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}