The fundamental frequency of a closed organ pipe of a length \(20\) cm is equal to the second overtone of an organ pipe open at both ends. The length of the organ pipe open at both ends will be:
1. | \(80\) cm | 2. | \(100\) cm |
3. | \(120\) cm | 4. | \(140\) cm |
If \(n_1\), \(n_2\), and \(n_3\) are the fundamental frequencies of three segments into which a string is divided, then the original fundamental frequency \(n\) of the string is given by:
1. \( \frac{1}{n}=\frac{1}{n_1}+\frac{1}{n_2}+\frac{1}{n_3}\)
2. \( \frac{1}{\sqrt{n}}=\frac{1}{\sqrt{n_1}}+\frac{1}{\sqrt{n_2}}+\frac{1}{\sqrt{n_3}}\)
3. \( \sqrt{n}=\sqrt{n_1}+\sqrt{n_2}+\sqrt{n_3}\)
4. \( n=n_1+n_2+n_3\)
The number of possible natural oscillations of the air column in a pipe closed at one end of length \(85\) cm whose frequencies lie below \(1250\) Hz are:(velocity of sound= \(340~\text{m/s}\)
1. \(4\)
2. \(5\)
3. \(7\)
4. \(6\)
1. | Odd harmonics of the fundamental frequency will be generated. |
2. | All harmonics of the fundamental frequency will be generated. |
3. | Pressure change will be maximum at both ends. |
4. | The open end will be an antinode. |