16:I[8216,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"BackButton"] 17:I[1553,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"QuestionViewer"] 15:T4e3,{"@context":"https://schema.org","@type":"Quiz","name":"Jee Advanced 2005 Waves PYQ with Solution","url":"https://pyq-test-series-platform-alsjdcxu2-sameers-projects-41c5a71e.vercel.app/jee-advanced/questions/2005/physics/waves/1lgdxs0a0","educationalLevel":"Entrance exam preparation","about":"Jee Advanced / Physics / Waves","hasPart":{"@type":"Question","name":"Jee Advanced 2005 Waves PYQ with Solution","text":"A transverse harmonic disturbance is produced in a string. The maximum transverse velocity is 3 m/s and the maximum transverse acceleration is 90 m/s^2 . If the wave velocity is 20 m/s, then find the waveform.","eduQuestionType":"mcq","acceptedAnswer":{"@type":"Answer","text":"C"},"comment":{"@type":"Comment","text":"Step 1: Understanding Symbols Let \\omega be the angular velocity and A be the amplitude. Step 2: Maximum Velocity The highest speed up and down (maximum transverse velocity) is V_{\\max} = \\omega A = 3~\\mathrm{m/s}. Step 3: Maximum Acceleration The highest acceleration up and down is \\omega^2 A = 90~\\mathrm{m/s}^2. Step 4: Wave Velocity Relation The speed that the wave itself travels is \\frac{\\omega}{k} = 20~\\mathrm{m/s}, where k is called the wave number. Step 5: Find Angular Velocity \\omega To..."}}}18:T550,

Step 1: Understanding Symbols

Let $$\omega$$ be the angular velocity and $$A$$ be the amplitude.

Step 2: Maximum Velocity

The highest speed up and down (maximum transverse velocity) is $$V_{\max} = \omega A = 3~\mathrm{m/s}$$.

Step 3: Maximu

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