Q)A wave moves with speed 300 m/s on a wire which is under the tension of 500N.Find how much tension must be changed to increase the speed to 312m/s.

My method: Since $v= \sqrt{T/μ}.....(i)$,where T is tension and μ the linear density of the rope,
$312 = \sqrt{T/μ}$
Dividing and solving, we get T=540.8 and therefore 40.8N Extra Force Must be provided.
BOOK METHOD: By differentiating both sides of velocity equation(i),:
$ \frac{dv}{dT}=\frac{1}{\sqrt{μT}}$
Dividing original equation(i) with differentiated one,we get,
dT= $(2T)\frac{dv}{v}$
Finally substituting the values $ T=500~, v= 300, ~dv=12$,we will get dT=40 N, which differs from the above answer by 2%. What is wrong with my method?


Your method is completely fine and it is the one that should be used. Let me write it down once more $$ v = \sqrt\frac{{T}}{{\mu}}$$ $$ 300 = \sqrt\frac{500}{\mu}$$ $$ \mu = \frac{500}{90000}$$ $$\mu = \frac{5}{900}$$ No we want $v=312$ so let's just put it in the equation which you have stated $$ 312 = \sqrt\frac{900T}{5}$$ $$ \frac{312\times 312\times 5}{900}= T$$ $$T= 540.8$$ So change in tension is $40.8~N$. The problem with the approach that your book has taekn is that, differentials are used for small changes, if you write $d$ before anything it means a very very small change (well how small is the topic of hyperreal numbers and I don't want to go into that). Your book has used $dv$ and $dT$ for very big changes and hence the answer that we have got is just an approximation.

Let me illustrate the flaw in a more easier way, let's imagine that we have to find the square root of $80$ provided that we know the square root of $81$ is $9$. We can go on like this $$ y = \sqrt{x}$$ $$ x_1=81 , ~ y_1 =9 ~~~~~~~~, x_2=80 ~, y_2=?$$ $$ \frac{dy}{dx} = \frac{1}{2\sqrt{x}}$$ $$ dy = \frac{dx}{2\sqrt{x}}$$ $$ dy = \frac{-1}{18} , ~~~~~ dy = -0.05555$$ $$ y_2 = y_1 +dy= 9 - 0.055555= 8.94445$$ So, the square root of $80$ is $8.944445$, but if we were to use the long division method or calculator the answer we would get is $\sqrt{80} =8.944427$ and if we try to calculate error then it is about $0.00020124$%. We have got such a good approximation because difference in our $x$ values is just 1. However, if we try to find the square root $64$ knowing only that $\sqrt{81}=9$, let's see what happens$$ x_1 = 81~~~, y_1=9 ~~~~~~~~~~~~~~ x_2=64 ,~~~y_2 =?$$ $$ \frac{dy}{dx} = \frac{1}{2\sqrt{x}}$$ $$ dy = \frac {dx}{2\sqrt{x}}$$ $$ dy = \frac{-17}{18} = -0.944444444$$ $$ y_2 = y_1 + dy = 9 - 0.9444444444 = 8.055555555$$ But as we know the square root of $64$ is $8$ and hence our error is about $0.69$% . Error increases the difference between two $x$ values increases. You can look over here for what I tried to explain.

Hope it helps.


Your last equation is for differentials, it is only exact when the differences ($dT,dv$) are infinitesimally small. For any differences that are finite the answer will be close but slightly off. What you did is essentially a Taylor expansion. If you write $T$ as function of $v$ (assuming $\mu$ is constant) you can use a value at known $v=v_0$ to get an approximate answer for values of $v$ that are near $v_0$. \begin{align}T(v)&=T(v_0)+\frac{dT}{dv}\cdot(v-v_0)+\mathcal{O}(v-v_0)^2\\ T(v)&=T(v_0)+\frac{dT}{dv}\Delta v+\mathcal{O}(\Delta v^2)\\ \implies\Delta T&=T(v)-T(v_0)\approx\frac{dT}{dv}\Delta v \end{align} Here $\mathcal{O(\Delta v^2})$ means a polynomial that depends on $\Delta v^2$ or higher powers of $\Delta v$. This means that if we use the approximation in the last equation, the error between this approximation and the exact answer will be of order $\Delta v^2$. This is actually great, since if we decrease $\Delta v$ the error will go down quadratically. If you repeat your calculation for multiple values of $\Delta v$ you would find that

  1. the approximation gets better and better as $\Delta v$ goes to zero
  2. The error (your approximation minus the exact answer) would look like a parabola as a function of $\Delta v$.

Note: the error is also proportional to $\Delta v^3,\Delta v^4$ etc. but if $\Delta v$ is small these terms will be neglible compared to $\Delta v^2$.

Note 2: Since $T=\mu v^2$ we indeed have $\frac{dT}{dv}=2\mu v=2T/v$

  • $\begingroup$ Your unstated conclusion is that the book method is wrong, I think. $\endgroup$ – Dr Chuck Nov 18 '19 at 12:11

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