We've been given an equation for a damped pendulum which is A = A0 x e^-kn Where A is the amplitude in cm, A0 is the initial amplitude, k is the damping constant and n is number of swings. I've collected results and plotted a graph of n against lnA - rearranging the equation gives lnA = -kn + lnA0, so k is the gradient of my graph. N has no units and ln(A/cm) has no units, so it looks like k wouldn't have units either, but I can't help but think that k would have units - I'm just not sure what they are. I thought it could be per second, but I'm far from certain and none of my physics teachers seem to know. Thanks in advance to anyone who can shed some light on this!
We've been given an equation for a damped pendulum which is A = A0 x e^-kn Where A is the amplitude in cm, A0 is the initial amplitude, k is the damping constant and n is number of swings. I've collected results and plotted a graph of n against lnA - rearranging the equation gives lnA = -kn + lnA0, so k is the gradient of my graph. N has no units and ln(A/cm) has no units, so it looks like k wouldn't have units either, but I can't help but think that k would have units - I'm just not sure what they are. I thought it could be per second, but I'm far from certain and none of my physics teachers seem to know. Thanks in advance to anyone who can shed some light on this!
We've been given an equation for a damped pendulum which is A = A0 x e^-kn Where A is the amplitude in cm, A0 is the initial amplitude, k is the damping constant and n is number of swings. I've collected results and plotted a graph of n against lnA - rearranging the equation gives lnA = -kn + lnA0, so k is the gradient of my graph. N has no units and ln(A/cm) has no units, so it looks like k wouldn't have units either, but I can't help but think that k would have units - I'm just not sure what they are. I thought it could be per second, but I'm far from certain and none of my physics teachers seem to know. Thanks in advance to anyone who can shed some light on this!
it's the log of something that has units and you don't specify, how will people looking at the graph understand it?
ln cm
people who write physics books are imo bad at mentioning this - they probably came up using log graph paper which looks like this rather than doing it in excel or whatever
it's the log of something that has units and you don't specify, how will people looking at the graph understand it?
ln cm
people who write physics books are imo bad at mentioning this - they probably came up using log graph paper which looks like this rather than doing it in excel or whatever
I'm sorry, I appreciate you trying to help me but I don't really understand what you mean! What haven't I specified? and yep, my physics book has virtually nothing on this sort of stuff!
it's the log of something that has units and you don't specify, how will people looking at the graph understand it?
ln cm
people who write physics books are imo bad at mentioning this - they probably came up using log graph paper which looks like this rather than doing it in excel or whatever
I'm not sure I understand your point - ln(A/cm) is actually the log of a dimensionless quantity - the argument is the ratio of the measurement (in cm) to 1 cm. This is routinely written as ln (A/cm) - I guess in principle it should be written as ln[A(cm)/1cm] but who can be bothered to do that?
We've been given an equation for a damped pendulum which is A = A0 x e^-kn Where A is the amplitude in cm, A0 is the initial amplitude, k is the damping constant and n is number of swings. I've collected results and plotted a graph of n against lnA - rearranging the equation gives lnA = -kn + lnA0, so k is the gradient of my graph. N has no units and ln(A/cm) has no units, so it looks like k wouldn't have units either, but I can't help but think that k would have units - I'm just not sure what they are. I thought it could be per second, but I'm far from certain and none of my physics teachers seem to know. Thanks in advance to anyone who can shed some light on this!
Damping constant or damping coefficient in physics usually relates to the damping force as it varies with the velocity of the moving object. What you have here is not usually referred to as the damping constant or coefficient, which, by the way could have units as determined by F = kv (Or F=kv2 possibly) F being the damping force and v the velocity.
But as I say, this is not what we have here.
In the equation you have the constant k is a ratio. It would be called the logarithmic decrement, and is a measure of how much the amplitude of each successive swing decreases as a fraction of the previous amplitude. As such this constant has no units and is dimensionless, as others have said here. It's a ratio. If you want to look at it more closely you can see that from
A=A0e−kn
You get
LnA0A=−kn
So if n is one oscillation, the value of k gives the log of the ratio of the amplitudes of two successive swings. To put some numbers in, if the amplitude of the 2nd swing was half the amplitude of the 1st Ln0.5=−k
Gives k = 0.69
So k= 0.69 is the value of your constant, for example, for the amplitude to be halved each swing. This is actually quite severe damping for a pendulum. Your experimental values for k will be much less. It has no unit. It needs no unit.
Wow, thank you so much for that very detailed reply that definitely helps, I was getting quite confused when I googled it because it kept coming up with information about the damping coefficient you mentioned instead, so I thought this constant would have to have a unit too! Thanks again
Thank you so much for this.In addition, could you please help me understand what the S.I unit of damping force would be?And if possible,The equation of damping co-efficient?Your response would be much appreciated!
Thank you so much for this.In addition, could you please help me understand what the S.I unit of damping force would be?And if possible,The equation of damping co-efficient?Your response would be much appreciated!
Hi, welcome to TSR. This thread had started about 5 years. If you have questions, you may want to start a new thread and make reference to this thread.
I would make a reply to this post but I would not reply for subsequent questions. If you still have questions, start a new thread.