JEE Physics - MCQ Practice Questions

When a wave reflects from a fixed (denser) end, it undergoes a phase change of π radians or 180°

From y = 0.02sin(50πt - 4πx), comparing with y = Asin(ωt - kx): ω = 50π rad/s, k = 4π rad/m. Wavelength λ = 2π/k = 2π/(4π) = 0.5 m. Wave velocity v = ω/k = 50π/(4π) = 12.5 m/s.

A resonance tube with water acts as a closed pipe at the water surface and open at the top, requiring the air column length to be an odd multiple of λ/4 for resonance, but the effective length includes end correction beyond the physical opening.

Step 1: Calculate Wavelength

[The wavelength of sound in air is found using the wave equation with given frequency and speed.]

Step 2: Determine Condition for First Resonance

[For a closed pipe (water surface acts as closed end), first resonance occurs at the shortest air column length, which is λ/4.]

Step 3: Account for End Correction

[The effective length of the air column is greater than the physical length because sound appears to resonate beyond the open end by approximately 1 cm due to end correction.]

Final Answer: (C) 17.2 cm; due to end correction of approximately 1 cm for the open end

The effective length accounts for the fact that the sound wave's effective vibrating column extends slightly beyond the physical opening of the tube at the open end.