Standing Wave in String and Organ Pipe: Practice Problem & Solution
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.
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Standing Wave in String and Organ Pipe. Understanding the underlying formula is key to arriving at the correct answer below:
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.
Leave a Reply