A tube closed at one end and containing air produces, when excited the fundamental note of frequency $512 \mathrm{~Hz}$. If the tube is open at both ends, the fundamental frequency that can be excited is (in Hz).
A. $128$
B. $256$
C. $512$
D. $1024$
Answer
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Hint: 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:
$v_{c}=\dfrac{v}{4 l}$
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
$v_{c}=\dfrac{v}{4 l}$
where variables have their usual meanings
Now, for a tube locked at one end, the basic frequency is
$v_{0}=\dfrac{v}{2 l}$
where variables have their usual meanings
$v_{\mathrm{o}}=\dfrac{1}{2} 512 \times 4$
$\therefore {{v}_{\text{o}}}=1024~\text{Hz}$
If the tube is open at both ends, the fundamental frequency that can be excited is $1024~\text{Hz}$
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.
Formula used:
$v_{c}=\dfrac{v}{4 l}$
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
$v_{c}=\dfrac{v}{4 l}$
where variables have their usual meanings
Now, for a tube locked at one end, the basic frequency is
$v_{0}=\dfrac{v}{2 l}$
where variables have their usual meanings
$v_{\mathrm{o}}=\dfrac{1}{2} 512 \times 4$
$\therefore {{v}_{\text{o}}}=1024~\text{Hz}$
If the tube is open at both ends, the fundamental frequency that can be excited is $1024~\text{Hz}$
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.
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