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Question: A tube closed at one end and containing air produces, when excited the fundamental note of frequency...

A tube closed at one end and containing air produces, when excited the fundamental note of frequency 512 Hz512 \mathrm{~Hz}. If the tube is open at both ends, the fundamental frequency that can be excited is (in Hz).
A. 128128
B. 256256
C. 512512
D. 10241024

Explanation

Solution

As the lowest frequency of a periodic waveform, the natural frequency, or fundamental frequency, often referred to simply as the fundamental, is defined. The fundamental thing in music is the musical pitch of a note that is perceived as the smallest partial present. Calculate the fundamental frequency of the tube and put it in the formula of the fundamental frequency.

Formula used:
vc=v4lv_{c}=\dfrac{v}{4 l}

Complete solution:
The tension and the mass per unit length of the string determine the velocity of a travelling wave in a stretched string. For a cm-length string and mass/length = gm/m. The fundamental frequency would be Hz for such a string.

The fundamental frequency for a tube closed at one end is
vc=v4lv_{c}=\dfrac{v}{4 l}
where variables have their usual meanings

Now, for a tube locked at one end, the basic frequency is
v0=v2lv_{0}=\dfrac{v}{2 l}
where variables have their usual meanings
vo=12512×4v_{\mathrm{o}}=\dfrac{1}{2} 512 \times 4
vo=1024 Hz\therefore {{v}_{\text{o}}}=1024~\text{Hz}

If the tube is open at both ends, the fundamental frequency that can be excited is 1024 Hz1024~\text{Hz}

Hence, the correct option is (D).

Note:
In any complex waveform, the fundamental frequency provides the sound with its strongest audible pitch reference - it is the predominant frequency. The simplest of all waveforms is a sine wave and contains only one basic frequency and no harmonics, overtones or partials. The following properties are described: When the string length is changed, it vibrates at a different frequency. Shorter strings have greater frequency and higher pitch as a result.