A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is 340 m s–1).

First (Fundamental) harmonic mode of the pipe will be resonantly excited by a 430 Hz source. The same source will not be in resonance with the pipe if both ends are open.


Explanation:


Given,


Length of the closed pipe, l = 20 cm = 0.2 m


Source frequency = nth normal mode of frequency, fn = 430 Hz


Speed of sound, v = 340 m s-1


In a closed pipe, the nth normal mode of frequency is given by



⇒


⇒ 2n-1 = 1.012


⇒ 2n = 2.012


⇒ n = 1.006 or 1


Hence, the frequency corresponding to the first (fundamental) mode of vibration is resonantly excited by the given source. In a pipe open at both ends, the nth mode of vibration frequency is given by


fn = nv/2l


⇒ n = 2lfn/v


NOTE: Resonance means that a certain frequency of wave hitting some object is "in synch" with the natural vibrating frequency of that object. If the wave is in synch, it reinforces the natural vibration of the object and can cause the amplitude (amount) of vibration of the object to increase greatly.


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