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Practice question

Question

A string vibrates with a stationary wave y = 0.1 sin (2Ï€ x/3) cos (150Ï€ t) . What is the distance between a node and the next antinode?

Options

Choose one · Correct answer highlighted

Explanation

Given: A string vibrates with a stationary wave y = 0.1 sin (2π x/3) cos (150π t) . What is the distance between a node and the next antinode? These values define the system as per NCERT data. Formula: k = 2π/3, lambda = 2π/k = 3 m. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Distance between node and antinode: lambda/4 = 3/4 = 0.75 m . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

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