Always appreciate your quality videos. In this instance, does the equation only hold if the angles are small, which was our assumption in the derivation. Thank you.
yes, however the analysis doesn't indicated how small. For large amplitudes there would be another term (non-linear wave equation) that would need to be included. For waves on a string the basic wave equation seems to do a pretty good job for larger amplitudes which tells me that the non-linear terms are probably small.
Why is delta(m) = mudelta(x)? Shouldn’t it be mudelta(s) which is the segment along the string? Or a we assuming since theta is small delta(s) and x are the same
At 5:17 -- I understand that you used the small angle approximation assuming that both theta1 and theta2 would be equal along line segments bounded by delta x. My question is how exactly were you able to use the small angle approximation for ALL points along the string. Take for instance the points between any trough and peak, how would you be able to apply the approximation to those points where theta1 and theta2 would (at least by first glance) not be small enough to having sin (theta) = tan (theta)?
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This uses something a little more involved than the small angle approximation, although it is roughly the same thing. Theta 1 and theta 2 are NOT the same angle, but differ by only a small quantity, let's call it a, so theta_1 = theta_2 + a. If we apply a Taylor series expansion on cos(theta_1) and cos(theta_2), which is by the way how these small angle approximations are done in the first place, then we find that the two differ by a sum involving a^2, a^3 and so on. Since a is small, all these high powers of a are absolutely negligible, so we can approximate the cosines as equal. This then implies, as in the video, that the tensions are equal. If we do the same thing comparing sin(theta_1) and sin(theta_2), then the two terms differ by a sum with leading term a. Thus, we say that to the FIRST ORDER (I.e. the first power of a, because a is small enough that any higher powers of a are too tiny to worry about), that this approximation is valid. This assumption is where the weird behaviour comes from- the cosines are similar enough that we can consider them equal, but the sines are still noticeably not the same angle
You can expand trig function and keep only terms up to linear terms. Others terms will be small if thetas are small. So cos theta is approx 1 and sin theta terms are approx theta.
That's just how physicists do their thing. I'm pretty sure this only works for an idealized model. A lot of the proofs I've seen come from idealized models, and the formulas that come out are pretty reliable.
Nice derivation. I have watched a few other derivations and yours seemed to make the most sense to me. Appreciate it. Thank you.
Glad it was helpful!
you explained one by one clearly. Thank you!
Always appreciate your quality videos. In this instance, does the equation only hold if the angles are small, which was our assumption in the derivation. Thank you.
yes, however the analysis doesn't indicated how small. For large amplitudes there would be another term (non-linear wave equation) that would need to be included. For waves on a string the basic wave equation seems to do a pretty good job for larger amplitudes which tells me that the non-linear terms are probably small.
elaborate explanation, thanks bro
You are welcome
Why is delta(m) = mudelta(x)?
Shouldn’t it be mudelta(s) which is the segment along the string?
Or a we assuming since theta is small delta(s) and x are the same
Yes
its bcoz mu(the mass density)= del m(mass)/del x(displacement)
At 5:17 -- I understand that you used the small angle approximation assuming that both theta1 and theta2 would be equal along line segments bounded by delta x. My question is how exactly were you able to use the small angle approximation for ALL points along the string. Take for instance the points between any trough and peak, how would you be able to apply the approximation to those points where theta1 and theta2 would (at least by first glance) not be small enough to having sin (theta) = tan (theta)?
You are making small angle approximation for 2 points that are infinitesimally close to each other.
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Since tension is unofirm and theta 1 equals theta 2 then sum of forces over Y equals zero not ma
This uses something a little more involved than the small angle approximation, although it is roughly the same thing. Theta 1 and theta 2 are NOT the same angle, but differ by only a small quantity, let's call it a, so theta_1 = theta_2 + a. If we apply a Taylor series expansion on cos(theta_1) and cos(theta_2), which is by the way how these small angle approximations are done in the first place, then we find that the two differ by a sum involving a^2, a^3 and so on. Since a is small, all these high powers of a are absolutely negligible, so we can approximate the cosines as equal. This then implies, as in the video, that the tensions are equal. If we do the same thing comparing sin(theta_1) and sin(theta_2), then the two terms differ by a sum with leading term a. Thus, we say that to the FIRST ORDER (I.e. the first power of a, because a is small enough that any higher powers of a are too tiny to worry about), that this approximation is valid. This assumption is where the weird behaviour comes from- the cosines are similar enough that we can consider them equal, but the sines are still noticeably not the same angle
What's the purpose of the small angle approximation here? And why do we assume theta is a relatively small angle?
To make the equation easier , you could find this for large angle but irl small angle works just fine
why do we use theta1=theta1 in the horizontal direction but not in the vertical direction?
You can expand trig function and keep only terms up to linear terms. Others terms will be small if thetas are small. So cos theta is approx 1 and sin theta terms are approx theta.
En büyük müsün bilmiyorum ama çok büyüksün ninja abi.
naah bro explained exponentially better than my proff. 😭
Будьласка допоможіть рівняння біжучої хвилі завдання на виведення з формули я нічого незнаю і не розумію.
Very helpful thank youuu
Thanks so much, really appreciate it🙌
Thank you! You are a beast
THANKS
You're welcome!
11:01 “previous” i didn’t watch. now i’m lost. please don’t tell me to watch another video now. 😢😢😢😢
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After this much approximation , I think the equation is literally useless.
That's just how physicists do their thing. I'm pretty sure this only works for an idealized model. A lot of the proofs I've seen come from idealized models, and the formulas that come out are pretty reliable.
welcome to physics
not bad
😂