question_answer

A tuning fork is in resonance with a closed pipe. But the same tuning fork cannot be in resonance with an open pipe of same length. Why?

Answer:

                As the tuning fork of frequency \[v\] is in resonance with a closed pipe of length \[L\], so \[v=\frac{\upsilon }{4L}\] But an open pipe of length L produces a frequency of \[\upsilon /2L\]. Hence it cannot be in resonance with the tuning fork of frequency \[v(=\upsilon /4L).\]