#Stationary Wave

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 t

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.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Oscillations and Waves, Topic: SHM, wave motion and superposition.