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?
Explanation
Given:
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?
These values define the system as per NCERT data.
Formula:
k = 2Ï€/3, lambda = 2Ï€/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 kgâ»Â¹ Kâ»Â¹, m/s², 10â»âµ are properly used as per NCERT.
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