Standing Waves
PHXI15:WAVES

354959 The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is \(60\,cm,\) the length of the closed pipe will be .

1 \(45\,cm\)
2 \(60\,cm\)
3 \(15\,cm\)
4 \(30\,cm\)
PHXI15:WAVES

354960 An open pipe and a closed pipe have same length. The ratio of \(p^{\text {th }}\) overtones of two pipes is

1 \(\dfrac{1}{p}\)
2 \(p\)
3 \(\dfrac{2(p+1)}{2 p+1}\)
4 \(\dfrac{2 p+1}{2(p+1)}\)
PHXI15:WAVES

354961 A cylindrical tube open at both the ends has a fundamental frequency of \(390\;Hz\) in air. If
\(1 / 4^{\text {th }}\) of the tube is immersed vertically in water the fundamental frequency of air column is

1 \(260\;Hz\)
2 \(130\;Hz\)
3 \(390\;Hz\)
4 \(520\;Hz\)
PHXI15:WAVES

354962 \(n_{1}\) is the frequency of the pipe closed at one end and \(n_{2}\) is the frequency of the pipe open at both ends. If both are joined end to end, find the fundamental frequency of closed pipe that is formed:

1 \(\dfrac{n_{1} n_{2}}{2 n_{2}+2 n_{1}}\)
2 \(\dfrac{n_{1} n_{2}}{n_{2}+2 n_{1}}\)
3 \(\dfrac{2 n_{1}+n_{2}}{n_{2} n_{1}}\)
4 \(\dfrac{n_{1}+2 n_{2}}{n_{2} 2 n_{1}}\)
PHXI15:WAVES

354959 The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is \(60\,cm,\) the length of the closed pipe will be .

1 \(45\,cm\)
2 \(60\,cm\)
3 \(15\,cm\)
4 \(30\,cm\)
PHXI15:WAVES

354960 An open pipe and a closed pipe have same length. The ratio of \(p^{\text {th }}\) overtones of two pipes is

1 \(\dfrac{1}{p}\)
2 \(p\)
3 \(\dfrac{2(p+1)}{2 p+1}\)
4 \(\dfrac{2 p+1}{2(p+1)}\)
PHXI15:WAVES

354961 A cylindrical tube open at both the ends has a fundamental frequency of \(390\;Hz\) in air. If
\(1 / 4^{\text {th }}\) of the tube is immersed vertically in water the fundamental frequency of air column is

1 \(260\;Hz\)
2 \(130\;Hz\)
3 \(390\;Hz\)
4 \(520\;Hz\)
PHXI15:WAVES

354962 \(n_{1}\) is the frequency of the pipe closed at one end and \(n_{2}\) is the frequency of the pipe open at both ends. If both are joined end to end, find the fundamental frequency of closed pipe that is formed:

1 \(\dfrac{n_{1} n_{2}}{2 n_{2}+2 n_{1}}\)
2 \(\dfrac{n_{1} n_{2}}{n_{2}+2 n_{1}}\)
3 \(\dfrac{2 n_{1}+n_{2}}{n_{2} n_{1}}\)
4 \(\dfrac{n_{1}+2 n_{2}}{n_{2} 2 n_{1}}\)
NEET Test Series from KOTA - 10 Papers In MS WORD WhatsApp Here
PHXI15:WAVES

354959 The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is \(60\,cm,\) the length of the closed pipe will be .

1 \(45\,cm\)
2 \(60\,cm\)
3 \(15\,cm\)
4 \(30\,cm\)
PHXI15:WAVES

354960 An open pipe and a closed pipe have same length. The ratio of \(p^{\text {th }}\) overtones of two pipes is

1 \(\dfrac{1}{p}\)
2 \(p\)
3 \(\dfrac{2(p+1)}{2 p+1}\)
4 \(\dfrac{2 p+1}{2(p+1)}\)
PHXI15:WAVES

354961 A cylindrical tube open at both the ends has a fundamental frequency of \(390\;Hz\) in air. If
\(1 / 4^{\text {th }}\) of the tube is immersed vertically in water the fundamental frequency of air column is

1 \(260\;Hz\)
2 \(130\;Hz\)
3 \(390\;Hz\)
4 \(520\;Hz\)
PHXI15:WAVES

354962 \(n_{1}\) is the frequency of the pipe closed at one end and \(n_{2}\) is the frequency of the pipe open at both ends. If both are joined end to end, find the fundamental frequency of closed pipe that is formed:

1 \(\dfrac{n_{1} n_{2}}{2 n_{2}+2 n_{1}}\)
2 \(\dfrac{n_{1} n_{2}}{n_{2}+2 n_{1}}\)
3 \(\dfrac{2 n_{1}+n_{2}}{n_{2} n_{1}}\)
4 \(\dfrac{n_{1}+2 n_{2}}{n_{2} 2 n_{1}}\)
PHXI15:WAVES

354959 The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is \(60\,cm,\) the length of the closed pipe will be .

1 \(45\,cm\)
2 \(60\,cm\)
3 \(15\,cm\)
4 \(30\,cm\)
PHXI15:WAVES

354960 An open pipe and a closed pipe have same length. The ratio of \(p^{\text {th }}\) overtones of two pipes is

1 \(\dfrac{1}{p}\)
2 \(p\)
3 \(\dfrac{2(p+1)}{2 p+1}\)
4 \(\dfrac{2 p+1}{2(p+1)}\)
PHXI15:WAVES

354961 A cylindrical tube open at both the ends has a fundamental frequency of \(390\;Hz\) in air. If
\(1 / 4^{\text {th }}\) of the tube is immersed vertically in water the fundamental frequency of air column is

1 \(260\;Hz\)
2 \(130\;Hz\)
3 \(390\;Hz\)
4 \(520\;Hz\)
PHXI15:WAVES

354962 \(n_{1}\) is the frequency of the pipe closed at one end and \(n_{2}\) is the frequency of the pipe open at both ends. If both are joined end to end, find the fundamental frequency of closed pipe that is formed:

1 \(\dfrac{n_{1} n_{2}}{2 n_{2}+2 n_{1}}\)
2 \(\dfrac{n_{1} n_{2}}{n_{2}+2 n_{1}}\)
3 \(\dfrac{2 n_{1}+n_{2}}{n_{2} n_{1}}\)
4 \(\dfrac{n_{1}+2 n_{2}}{n_{2} 2 n_{1}}\)