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Are sinusoidal travelling waves also normal modes of vibration?
Questions related to resonance/standing-waves and soundWhat's the physical interpretation of an arbitrary normal mode for masses and springs?In a lattice, what, technically, are random (thermal) atomic motions?Why doesn't a stationary wave completely destruct, instead of just at the nodes, when the waves are in fixed positions (see examples)Traveling wave solutions for an irregular “waveguide”Standing waves due to two counter-propagating travelling waves of different amplitudeEnergy Conservation of waves at a boundaryAt what frequency does a string vibrate?Why do musicians stretch the strings of their string instruments?Shouldn't reflection at the boundary interfere with the original wave to not give any wave?
$begingroup$
According to definition of normal modes, which says if all the different independent parts of a system vibrate at same frequency and their amplitude preserve a fixed ratio then such a motion is a normal mode of that system then since in sinusoidal travelling waves also different parts move with same frequency and different parts preserve a ratio, shouldn't they too be normal modes?
So are sinusoidal traveling waves normal modes?
newtonian-mechanics waves vibrations
asked 4 hours ago
LuciferLucifer
245
$endgroup$
add a comment |
$begingroup$
According to definition of normal modes, which says if all the different independent parts of a system vibrate at same frequency and their amplitude preserve a fixed ratio then such a motion is a normal mode of that system then since in sinusoidal travelling waves also different parts move with same frequency and different parts preserve a ratio, shouldn't they too be normal modes?
So are sinusoidal traveling waves normal modes?
newtonian-mechanics waves vibrations
asked 4 hours ago
LuciferLucifer
245
$endgroup$
add a comment |
$begingroup$
According to definition of normal modes, which says if all the different independent parts of a system vibrate at same frequency and their amplitude preserve a fixed ratio then such a motion is a normal mode of that system then since in sinusoidal travelling waves also different parts move with same frequency and different parts preserve a ratio, shouldn't they too be normal modes?
So are sinusoidal traveling waves normal modes?
newtonian-mechanics waves vibrations
asked 4 hours ago
LuciferLucifer
245
$endgroup$
According to definition of normal modes, which says if all the different independent parts of a system vibrate at same frequency and their amplitude preserve a fixed ratio then such a motion is a normal mode of that system then since in sinusoidal travelling waves also different parts move with same frequency and different parts preserve a ratio, shouldn't they too be normal modes?
So are sinusoidal traveling waves normal modes?
newtonian-mechanics waves vibrations
newtonian-mechanics waves vibrations
asked 4 hours ago
LuciferLucifer
245
asked 4 hours ago
LuciferLucifer
245
asked 4 hours ago
LuciferLucifer
245
asked 4 hours ago
LuciferLucifer
245
asked 4 hours ago
LuciferLucifer
245
245
add a comment |
add a comment |
1 Answer
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If the equations of motion of the vibrating system are equivalent to real and symmetric mass and stiffness terms, the normal modes will be real vectors, which means that all parts of the system move in the same phase.
That excludes travelling waves, where there is a phase difference between points in the direction of travel of the wave.
There is a special case, when two (or more) vibration modes have identical frequencies. In that situation, a combination of the different mode shapes with different phases may "look like" a travelling wave. However this may only be a theoretical possibility, because the tolerances in a real-life structures often separate the two theoretically-identical frequencies.
However there are mechanical systems which do have "travelling" normal vibration modes. A simple example is a gyroscope, where the vibration modes include precession and nutation.
In general, the equations of motion of a system rotating with constant angular velocity will include Coriolis terms, if it is modelled in a rotating coordinate system fixed to the undeformed shape of the body. The equations of motion are then Hermitian matrices rather than real symmetric matrices. The eigenvalues (natural frequencies) are still real, but the mode shapes are now complex vectors which can be interpreted as travelling waves.
In general, the speed at which the "mode shape" rotates around the object is different from the rotation speed of the object itself. If the two speeds coincide for some particular rotation speeds, that can have severe consequences for the design of real rotating machinery - for example the so-called "critical speeds" of rotating shafts.
answered 4 hours ago
alephzeroalephzero
5,53621120
$endgroup$
$begingroup$
"That excludes travelling waves, where there is a phase difference between points in the direction of travel of the wave."- but even in a standing wave, which is a normal mode for a string fixed at both ends, there is phase difference between points separated by a node or in two adjacent "loops". Isn't the requirement of a normal mode is that all moving parts have same frequency and not necessarily same phase?
$endgroup$
– Lucifer
3 hours ago
$begingroup$
Tell me if I am wrong, but I think that the reason sinusoidal travelling wave is not a normal mode because the ratio of amplitude of different parts don't remain constant. Like say the ratio between amplitude at two points A and B is say k but as time passes there will be a time when the same ratio is (1/k) or may be some other value which doesn't happen in case of normal modes like standing waves on a string.
$endgroup$
– Lucifer
3 hours ago
add a comment |
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1 Answer
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$begingroup$
If the equations of motion of the vibrating system are equivalent to real and symmetric mass and stiffness terms, the normal modes will be real vectors, which means that all parts of the system move in the same phase.
That excludes travelling waves, where there is a phase difference between points in the direction of travel of the wave.
There is a special case, when two (or more) vibration modes have identical frequencies. In that situation, a combination of the different mode shapes with different phases may "look like" a travelling wave. However this may only be a theoretical possibility, because the tolerances in a real-life structures often separate the two theoretically-identical frequencies.
However there are mechanical systems which do have "travelling" normal vibration modes. A simple example is a gyroscope, where the vibration modes include precession and nutation.
In general, the equations of motion of a system rotating with constant angular velocity will include Coriolis terms, if it is modelled in a rotating coordinate system fixed to the undeformed shape of the body. The equations of motion are then Hermitian matrices rather than real symmetric matrices. The eigenvalues (natural frequencies) are still real, but the mode shapes are now complex vectors which can be interpreted as travelling waves.
In general, the speed at which the "mode shape" rotates around the object is different from the rotation speed of the object itself. If the two speeds coincide for some particular rotation speeds, that can have severe consequences for the design of real rotating machinery - for example the so-called "critical speeds" of rotating shafts.
answered 4 hours ago
alephzeroalephzero
5,53621120
$endgroup$
$begingroup$
"That excludes travelling waves, where there is a phase difference between points in the direction of travel of the wave."- but even in a standing wave, which is a normal mode for a string fixed at both ends, there is phase difference between points separated by a node or in two adjacent "loops". Isn't the requirement of a normal mode is that all moving parts have same frequency and not necessarily same phase?
$endgroup$
– Lucifer
3 hours ago
$begingroup$
Tell me if I am wrong, but I think that the reason sinusoidal travelling wave is not a normal mode because the ratio of amplitude of different parts don't remain constant. Like say the ratio between amplitude at two points A and B is say k but as time passes there will be a time when the same ratio is (1/k) or may be some other value which doesn't happen in case of normal modes like standing waves on a string.
$endgroup$
– Lucifer
3 hours ago
add a comment |
$begingroup$
If the equations of motion of the vibrating system are equivalent to real and symmetric mass and stiffness terms, the normal modes will be real vectors, which means that all parts of the system move in the same phase.
That excludes travelling waves, where there is a phase difference between points in the direction of travel of the wave.
There is a special case, when two (or more) vibration modes have identical frequencies. In that situation, a combination of the different mode shapes with different phases may "look like" a travelling wave. However this may only be a theoretical possibility, because the tolerances in a real-life structures often separate the two theoretically-identical frequencies.
However there are mechanical systems which do have "travelling" normal vibration modes. A simple example is a gyroscope, where the vibration modes include precession and nutation.
In general, the equations of motion of a system rotating with constant angular velocity will include Coriolis terms, if it is modelled in a rotating coordinate system fixed to the undeformed shape of the body. The equations of motion are then Hermitian matrices rather than real symmetric matrices. The eigenvalues (natural frequencies) are still real, but the mode shapes are now complex vectors which can be interpreted as travelling waves.
In general, the speed at which the "mode shape" rotates around the object is different from the rotation speed of the object itself. If the two speeds coincide for some particular rotation speeds, that can have severe consequences for the design of real rotating machinery - for example the so-called "critical speeds" of rotating shafts.
answered 4 hours ago
alephzeroalephzero
5,53621120
$endgroup$
$begingroup$
"That excludes travelling waves, where there is a phase difference between points in the direction of travel of the wave."- but even in a standing wave, which is a normal mode for a string fixed at both ends, there is phase difference between points separated by a node or in two adjacent "loops". Isn't the requirement of a normal mode is that all moving parts have same frequency and not necessarily same phase?
$endgroup$
– Lucifer
3 hours ago
$begingroup$
Tell me if I am wrong, but I think that the reason sinusoidal travelling wave is not a normal mode because the ratio of amplitude of different parts don't remain constant. Like say the ratio between amplitude at two points A and B is say k but as time passes there will be a time when the same ratio is (1/k) or may be some other value which doesn't happen in case of normal modes like standing waves on a string.
$endgroup$
– Lucifer
3 hours ago
add a comment |
$begingroup$
If the equations of motion of the vibrating system are equivalent to real and symmetric mass and stiffness terms, the normal modes will be real vectors, which means that all parts of the system move in the same phase.
That excludes travelling waves, where there is a phase difference between points in the direction of travel of the wave.
There is a special case, when two (or more) vibration modes have identical frequencies. In that situation, a combination of the different mode shapes with different phases may "look like" a travelling wave. However this may only be a theoretical possibility, because the tolerances in a real-life structures often separate the two theoretically-identical frequencies.
However there are mechanical systems which do have "travelling" normal vibration modes. A simple example is a gyroscope, where the vibration modes include precession and nutation.
In general, the equations of motion of a system rotating with constant angular velocity will include Coriolis terms, if it is modelled in a rotating coordinate system fixed to the undeformed shape of the body. The equations of motion are then Hermitian matrices rather than real symmetric matrices. The eigenvalues (natural frequencies) are still real, but the mode shapes are now complex vectors which can be interpreted as travelling waves.
In general, the speed at which the "mode shape" rotates around the object is different from the rotation speed of the object itself. If the two speeds coincide for some particular rotation speeds, that can have severe consequences for the design of real rotating machinery - for example the so-called "critical speeds" of rotating shafts.
answered 4 hours ago
alephzeroalephzero
5,53621120
$endgroup$
If the equations of motion of the vibrating system are equivalent to real and symmetric mass and stiffness terms, the normal modes will be real vectors, which means that all parts of the system move in the same phase.
That excludes travelling waves, where there is a phase difference between points in the direction of travel of the wave.
There is a special case, when two (or more) vibration modes have identical frequencies. In that situation, a combination of the different mode shapes with different phases may "look like" a travelling wave. However this may only be a theoretical possibility, because the tolerances in a real-life structures often separate the two theoretically-identical frequencies.
However there are mechanical systems which do have "travelling" normal vibration modes. A simple example is a gyroscope, where the vibration modes include precession and nutation.
In general, the equations of motion of a system rotating with constant angular velocity will include Coriolis terms, if it is modelled in a rotating coordinate system fixed to the undeformed shape of the body. The equations of motion are then Hermitian matrices rather than real symmetric matrices. The eigenvalues (natural frequencies) are still real, but the mode shapes are now complex vectors which can be interpreted as travelling waves.
In general, the speed at which the "mode shape" rotates around the object is different from the rotation speed of the object itself. If the two speeds coincide for some particular rotation speeds, that can have severe consequences for the design of real rotating machinery - for example the so-called "critical speeds" of rotating shafts.
answered 4 hours ago
alephzeroalephzero
5,53621120
edited 3 hours ago
answered 4 hours ago
alephzeroalephzero
5,53621120
answered 4 hours ago
alephzeroalephzero
5,53621120
answered 4 hours ago
alephzeroalephzero
5,53621120
5,53621120
$begingroup$
"That excludes travelling waves, where there is a phase difference between points in the direction of travel of the wave."- but even in a standing wave, which is a normal mode for a string fixed at both ends, there is phase difference between points separated by a node or in two adjacent "loops". Isn't the requirement of a normal mode is that all moving parts have same frequency and not necessarily same phase?
$endgroup$
– Lucifer
3 hours ago
$begingroup$
Tell me if I am wrong, but I think that the reason sinusoidal travelling wave is not a normal mode because the ratio of amplitude of different parts don't remain constant. Like say the ratio between amplitude at two points A and B is say k but as time passes there will be a time when the same ratio is (1/k) or may be some other value which doesn't happen in case of normal modes like standing waves on a string.
$endgroup$
– Lucifer
3 hours ago
add a comment |
$begingroup$
"That excludes travelling waves, where there is a phase difference between points in the direction of travel of the wave."- but even in a standing wave, which is a normal mode for a string fixed at both ends, there is phase difference between points separated by a node or in two adjacent "loops". Isn't the requirement of a normal mode is that all moving parts have same frequency and not necessarily same phase?
$endgroup$
– Lucifer
3 hours ago
$begingroup$
Tell me if I am wrong, but I think that the reason sinusoidal travelling wave is not a normal mode because the ratio of amplitude of different parts don't remain constant. Like say the ratio between amplitude at two points A and B is say k but as time passes there will be a time when the same ratio is (1/k) or may be some other value which doesn't happen in case of normal modes like standing waves on a string.
$endgroup$
– Lucifer
3 hours ago
$begingroup$
"That excludes travelling waves, where there is a phase difference between points in the direction of travel of the wave."- but even in a standing wave, which is a normal mode for a string fixed at both ends, there is phase difference between points separated by a node or in two adjacent "loops". Isn't the requirement of a normal mode is that all moving parts have same frequency and not necessarily same phase?
$endgroup$
– Lucifer
3 hours ago
$begingroup$
"That excludes travelling waves, where there is a phase difference between points in the direction of travel of the wave."- but even in a standing wave, which is a normal mode for a string fixed at both ends, there is phase difference between points separated by a node or in two adjacent "loops". Isn't the requirement of a normal mode is that all moving parts have same frequency and not necessarily same phase?
$endgroup$
– Lucifer
3 hours ago
$begingroup$
Tell me if I am wrong, but I think that the reason sinusoidal travelling wave is not a normal mode because the ratio of amplitude of different parts don't remain constant. Like say the ratio between amplitude at two points A and B is say k but as time passes there will be a time when the same ratio is (1/k) or may be some other value which doesn't happen in case of normal modes like standing waves on a string.
$endgroup$
– Lucifer
3 hours ago
$begingroup$
Tell me if I am wrong, but I think that the reason sinusoidal travelling wave is not a normal mode because the ratio of amplitude of different parts don't remain constant. Like say the ratio between amplitude at two points A and B is say k but as time passes there will be a time when the same ratio is (1/k) or may be some other value which doesn't happen in case of normal modes like standing waves on a string.
$endgroup$
– Lucifer
3 hours ago
add a comment |
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