354912
An open pipe is suddenly closed at one end. As a result the frequency of third harmonic of the closed pipe is found to be higher by \(100\;Hz\). The fundamental frequency of open pipe is
1 \(30\;Hz\)
2 \(200\;Hz\)
3 \(480\;Hz\)
4 \(240\;Hz\)
Explanation:
\(f_{0 \text { (closed })}=\dfrac{v}{\lambda}=\dfrac{v}{4 l}\) third harmonic of closed pipe \(=3 f_{0 \text { (closed) }}=\dfrac{3 v}{4 l}\) \(\dfrac{3 v}{4 l}-\dfrac{v}{2 l}=100\) or \(\dfrac{v}{4 l}=100\) and \(f_{0(\text { open })}=\dfrac{v}{2 l} \Rightarrow \dfrac{v}{2 l}=200 H z\).
PHXI15:WAVES
354913
Velocity of sound in air is \(320\;m{s^{ - 1}}\). A pipe closed at one end has a length of \(1\;m\). Neglecting end correction, the air column in the pipe cannot resonate with sound of frequency
1 \(240\;Hz\)
2 \(80\;Hz\)
3 \(400\;Hz\)
4 \(320\;Hz\)
Explanation:
In a closed organ pipe the fundamental frequency is \(f = \frac{v}{{4L}}f = \frac{{320\;m{s^{ - 1}}}}{{4 \times 1\;{m^2}}} = 80\;Hz\) In a closed organ pipe only odd harmonics are present. So, it can resonate with \(80\;Hz,240\;Hz,{\text{ }}400\;Hz,560\;Hz\). Option (4) is correct.
PHXI15:WAVES
354914
A sufficiently long closed organ pipe has a small hole at its bottom. Initially, the pipe is empty. Water is poured into the pipe at a constant rate. The fundamental frequency of the air column in the pipe
1 Continuously increase
2 First increases and then becomes constant
3 Continuously decreases
4 First decreases and then becomes constant
Explanation:
Fundamental frequency \(f\) is \(f=\dfrac{v}{4 \ell}\) when the pipe gets filled up \(\ell\) decreases and hence \(f\) increases. When the inflow and outflow of water are equal \(\ell\) becomes constant and hence \(f\) also becomes constant.
PHXI15:WAVES
354915
A closed pipe has certain frequency. Now its length is halved. Considering the end correction, its frequency will now become
354912
An open pipe is suddenly closed at one end. As a result the frequency of third harmonic of the closed pipe is found to be higher by \(100\;Hz\). The fundamental frequency of open pipe is
1 \(30\;Hz\)
2 \(200\;Hz\)
3 \(480\;Hz\)
4 \(240\;Hz\)
Explanation:
\(f_{0 \text { (closed })}=\dfrac{v}{\lambda}=\dfrac{v}{4 l}\) third harmonic of closed pipe \(=3 f_{0 \text { (closed) }}=\dfrac{3 v}{4 l}\) \(\dfrac{3 v}{4 l}-\dfrac{v}{2 l}=100\) or \(\dfrac{v}{4 l}=100\) and \(f_{0(\text { open })}=\dfrac{v}{2 l} \Rightarrow \dfrac{v}{2 l}=200 H z\).
PHXI15:WAVES
354913
Velocity of sound in air is \(320\;m{s^{ - 1}}\). A pipe closed at one end has a length of \(1\;m\). Neglecting end correction, the air column in the pipe cannot resonate with sound of frequency
1 \(240\;Hz\)
2 \(80\;Hz\)
3 \(400\;Hz\)
4 \(320\;Hz\)
Explanation:
In a closed organ pipe the fundamental frequency is \(f = \frac{v}{{4L}}f = \frac{{320\;m{s^{ - 1}}}}{{4 \times 1\;{m^2}}} = 80\;Hz\) In a closed organ pipe only odd harmonics are present. So, it can resonate with \(80\;Hz,240\;Hz,{\text{ }}400\;Hz,560\;Hz\). Option (4) is correct.
PHXI15:WAVES
354914
A sufficiently long closed organ pipe has a small hole at its bottom. Initially, the pipe is empty. Water is poured into the pipe at a constant rate. The fundamental frequency of the air column in the pipe
1 Continuously increase
2 First increases and then becomes constant
3 Continuously decreases
4 First decreases and then becomes constant
Explanation:
Fundamental frequency \(f\) is \(f=\dfrac{v}{4 \ell}\) when the pipe gets filled up \(\ell\) decreases and hence \(f\) increases. When the inflow and outflow of water are equal \(\ell\) becomes constant and hence \(f\) also becomes constant.
PHXI15:WAVES
354915
A closed pipe has certain frequency. Now its length is halved. Considering the end correction, its frequency will now become
354912
An open pipe is suddenly closed at one end. As a result the frequency of third harmonic of the closed pipe is found to be higher by \(100\;Hz\). The fundamental frequency of open pipe is
1 \(30\;Hz\)
2 \(200\;Hz\)
3 \(480\;Hz\)
4 \(240\;Hz\)
Explanation:
\(f_{0 \text { (closed })}=\dfrac{v}{\lambda}=\dfrac{v}{4 l}\) third harmonic of closed pipe \(=3 f_{0 \text { (closed) }}=\dfrac{3 v}{4 l}\) \(\dfrac{3 v}{4 l}-\dfrac{v}{2 l}=100\) or \(\dfrac{v}{4 l}=100\) and \(f_{0(\text { open })}=\dfrac{v}{2 l} \Rightarrow \dfrac{v}{2 l}=200 H z\).
PHXI15:WAVES
354913
Velocity of sound in air is \(320\;m{s^{ - 1}}\). A pipe closed at one end has a length of \(1\;m\). Neglecting end correction, the air column in the pipe cannot resonate with sound of frequency
1 \(240\;Hz\)
2 \(80\;Hz\)
3 \(400\;Hz\)
4 \(320\;Hz\)
Explanation:
In a closed organ pipe the fundamental frequency is \(f = \frac{v}{{4L}}f = \frac{{320\;m{s^{ - 1}}}}{{4 \times 1\;{m^2}}} = 80\;Hz\) In a closed organ pipe only odd harmonics are present. So, it can resonate with \(80\;Hz,240\;Hz,{\text{ }}400\;Hz,560\;Hz\). Option (4) is correct.
PHXI15:WAVES
354914
A sufficiently long closed organ pipe has a small hole at its bottom. Initially, the pipe is empty. Water is poured into the pipe at a constant rate. The fundamental frequency of the air column in the pipe
1 Continuously increase
2 First increases and then becomes constant
3 Continuously decreases
4 First decreases and then becomes constant
Explanation:
Fundamental frequency \(f\) is \(f=\dfrac{v}{4 \ell}\) when the pipe gets filled up \(\ell\) decreases and hence \(f\) increases. When the inflow and outflow of water are equal \(\ell\) becomes constant and hence \(f\) also becomes constant.
PHXI15:WAVES
354915
A closed pipe has certain frequency. Now its length is halved. Considering the end correction, its frequency will now become
354912
An open pipe is suddenly closed at one end. As a result the frequency of third harmonic of the closed pipe is found to be higher by \(100\;Hz\). The fundamental frequency of open pipe is
1 \(30\;Hz\)
2 \(200\;Hz\)
3 \(480\;Hz\)
4 \(240\;Hz\)
Explanation:
\(f_{0 \text { (closed })}=\dfrac{v}{\lambda}=\dfrac{v}{4 l}\) third harmonic of closed pipe \(=3 f_{0 \text { (closed) }}=\dfrac{3 v}{4 l}\) \(\dfrac{3 v}{4 l}-\dfrac{v}{2 l}=100\) or \(\dfrac{v}{4 l}=100\) and \(f_{0(\text { open })}=\dfrac{v}{2 l} \Rightarrow \dfrac{v}{2 l}=200 H z\).
PHXI15:WAVES
354913
Velocity of sound in air is \(320\;m{s^{ - 1}}\). A pipe closed at one end has a length of \(1\;m\). Neglecting end correction, the air column in the pipe cannot resonate with sound of frequency
1 \(240\;Hz\)
2 \(80\;Hz\)
3 \(400\;Hz\)
4 \(320\;Hz\)
Explanation:
In a closed organ pipe the fundamental frequency is \(f = \frac{v}{{4L}}f = \frac{{320\;m{s^{ - 1}}}}{{4 \times 1\;{m^2}}} = 80\;Hz\) In a closed organ pipe only odd harmonics are present. So, it can resonate with \(80\;Hz,240\;Hz,{\text{ }}400\;Hz,560\;Hz\). Option (4) is correct.
PHXI15:WAVES
354914
A sufficiently long closed organ pipe has a small hole at its bottom. Initially, the pipe is empty. Water is poured into the pipe at a constant rate. The fundamental frequency of the air column in the pipe
1 Continuously increase
2 First increases and then becomes constant
3 Continuously decreases
4 First decreases and then becomes constant
Explanation:
Fundamental frequency \(f\) is \(f=\dfrac{v}{4 \ell}\) when the pipe gets filled up \(\ell\) decreases and hence \(f\) increases. When the inflow and outflow of water are equal \(\ell\) becomes constant and hence \(f\) also becomes constant.
PHXI15:WAVES
354915
A closed pipe has certain frequency. Now its length is halved. Considering the end correction, its frequency will now become
