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An open pipe is in resonance in its 2nd harmonic with tuning fork of frequency f_(1).Now it is closed at one end. If the frequency of the tuning fork is increased slowly from f_(1)then again a resonance is obtained with a frequency f_(2) . If in this case the pipe vibrates nth harmonics then |
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Answer» `n=3, f_(2) = 3/4 f_(1)`  `therefore f_(1)`, the second harmonic of `f_(1) =(2V)/(2l) = v/l` , If one end is closed, it gives only odd harmonics. If fundamental `=v//4l`, The harmonic are `(3v)/(4l), (5v)/(4l)` etc. once, the frequency starts increasing, the FIRST higher harmonic that is resonated `=(3v)/(4l)` If n=3, `f_(2) =3/4.v/l = 3/4 f_(1)` However, here is a snag. The frequency increased from v/l. `therefore 3/4f_(1)` is not greater than `f_(1)`. `therefore 5/4 f_(1)` is the answer because this is greater than `f_(1)`. So, the ansser is (b) |
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