NEET Physics Question - Waves - Standing Wave in String and Organ Pipe

Assertion (A): A tuning fork is in resonance with a closed pipe in fundamental mode, but the same tuning fork cannot be in resonance in fundamental mode with an open pipe of same length.

Reason (R): The same tuning fork will not be in resonance with open pipe of same length due to end correction of pipe.

Both (A) & (R) are true and the (R) is the correct explanation of the (A)

Both (A) & (R) are true but the (R) is not the correct explanation of the (A)

(A) is true but (R) is false

Both (A) and (R) are false

Solution:

Assertion (A) is true. For a closed pipe of length \(L\), fundamental frequency is \(v/(4L)\). For an open pipe of same length, it's \(v/(2L)\). These are different, so the same tuning fork cannot resonate in fundamental mode with both. Reason (R) is false; the primary reason is the fundamental frequency difference (\(f_o = 2f_c\)), not solely end correction.

Rankers Physics

Resonance in Organ Pipes

Solution:

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