354944
The fundamental frequency of an air column in a pipe closed at one end is \(100\;Hz\). If the same pipe is open at both the ends, the frequencies produced in \(Hz\) are
1 \(100,200,300,400 \ldots \ldots\)
2 \(100,300,500,700 \ldots \ldots\)
3 \(200,300,400,500 \ldots \ldots\)
4 \(200,400,600,800 \ldots\)
Explanation:
For a closed pipe fundamental frequency, \({v_1} = \frac{v}{{4L}} = 100\;Hz\) For an open pipe fundamental frequency \({v_1} = \frac{v}{{2L}} = 200\;Hz\) In a pipe open at both the ends, all multiples of the fundamental are produced. So option (4) is correct
MHTCET - 2017
PHXI15:WAVES
354945
A cylindrical tube open at both ends has a fundamental frequency \(n\) in air. The tube is dipped vertically in water so that half of it is immersed in water. The fundamental frequency of air column is
1 \(n/2\)
2 \(n\)
3 \(2\,n\)
4 \(4\,n\)
Explanation:
When tube is open, \(n = \frac{v}{{2\ell }}\), where \(n\) is fundamental frequency of open organ pipe in air. When half of tube is dipped vertically in water, it behaves as a closed pipe of length \(\dfrac{\ell}{2}\), so fundamental frequency \(n\) in this case is \(\therefore \;\;\;{\mkern 1mu} {\kern 1pt} {n^\prime } = \frac{v}{{4(\ell /2)}} = \frac{v}{{2\ell }} = n\)
354944
The fundamental frequency of an air column in a pipe closed at one end is \(100\;Hz\). If the same pipe is open at both the ends, the frequencies produced in \(Hz\) are
1 \(100,200,300,400 \ldots \ldots\)
2 \(100,300,500,700 \ldots \ldots\)
3 \(200,300,400,500 \ldots \ldots\)
4 \(200,400,600,800 \ldots\)
Explanation:
For a closed pipe fundamental frequency, \({v_1} = \frac{v}{{4L}} = 100\;Hz\) For an open pipe fundamental frequency \({v_1} = \frac{v}{{2L}} = 200\;Hz\) In a pipe open at both the ends, all multiples of the fundamental are produced. So option (4) is correct
MHTCET - 2017
PHXI15:WAVES
354945
A cylindrical tube open at both ends has a fundamental frequency \(n\) in air. The tube is dipped vertically in water so that half of it is immersed in water. The fundamental frequency of air column is
1 \(n/2\)
2 \(n\)
3 \(2\,n\)
4 \(4\,n\)
Explanation:
When tube is open, \(n = \frac{v}{{2\ell }}\), where \(n\) is fundamental frequency of open organ pipe in air. When half of tube is dipped vertically in water, it behaves as a closed pipe of length \(\dfrac{\ell}{2}\), so fundamental frequency \(n\) in this case is \(\therefore \;\;\;{\mkern 1mu} {\kern 1pt} {n^\prime } = \frac{v}{{4(\ell /2)}} = \frac{v}{{2\ell }} = n\)
354944
The fundamental frequency of an air column in a pipe closed at one end is \(100\;Hz\). If the same pipe is open at both the ends, the frequencies produced in \(Hz\) are
1 \(100,200,300,400 \ldots \ldots\)
2 \(100,300,500,700 \ldots \ldots\)
3 \(200,300,400,500 \ldots \ldots\)
4 \(200,400,600,800 \ldots\)
Explanation:
For a closed pipe fundamental frequency, \({v_1} = \frac{v}{{4L}} = 100\;Hz\) For an open pipe fundamental frequency \({v_1} = \frac{v}{{2L}} = 200\;Hz\) In a pipe open at both the ends, all multiples of the fundamental are produced. So option (4) is correct
MHTCET - 2017
PHXI15:WAVES
354945
A cylindrical tube open at both ends has a fundamental frequency \(n\) in air. The tube is dipped vertically in water so that half of it is immersed in water. The fundamental frequency of air column is
1 \(n/2\)
2 \(n\)
3 \(2\,n\)
4 \(4\,n\)
Explanation:
When tube is open, \(n = \frac{v}{{2\ell }}\), where \(n\) is fundamental frequency of open organ pipe in air. When half of tube is dipped vertically in water, it behaves as a closed pipe of length \(\dfrac{\ell}{2}\), so fundamental frequency \(n\) in this case is \(\therefore \;\;\;{\mkern 1mu} {\kern 1pt} {n^\prime } = \frac{v}{{4(\ell /2)}} = \frac{v}{{2\ell }} = n\)
354944
The fundamental frequency of an air column in a pipe closed at one end is \(100\;Hz\). If the same pipe is open at both the ends, the frequencies produced in \(Hz\) are
1 \(100,200,300,400 \ldots \ldots\)
2 \(100,300,500,700 \ldots \ldots\)
3 \(200,300,400,500 \ldots \ldots\)
4 \(200,400,600,800 \ldots\)
Explanation:
For a closed pipe fundamental frequency, \({v_1} = \frac{v}{{4L}} = 100\;Hz\) For an open pipe fundamental frequency \({v_1} = \frac{v}{{2L}} = 200\;Hz\) In a pipe open at both the ends, all multiples of the fundamental are produced. So option (4) is correct
MHTCET - 2017
PHXI15:WAVES
354945
A cylindrical tube open at both ends has a fundamental frequency \(n\) in air. The tube is dipped vertically in water so that half of it is immersed in water. The fundamental frequency of air column is
1 \(n/2\)
2 \(n\)
3 \(2\,n\)
4 \(4\,n\)
Explanation:
When tube is open, \(n = \frac{v}{{2\ell }}\), where \(n\) is fundamental frequency of open organ pipe in air. When half of tube is dipped vertically in water, it behaves as a closed pipe of length \(\dfrac{\ell}{2}\), so fundamental frequency \(n\) in this case is \(\therefore \;\;\;{\mkern 1mu} {\kern 1pt} {n^\prime } = \frac{v}{{4(\ell /2)}} = \frac{v}{{2\ell }} = n\)
