# Why does my answer vary?

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,
$$300=\sqrt{500/μ}$$
$$312 = \sqrt{T/μ}$$
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,
$$\frac{dv}{v}=\frac{dT}{2T}$$,
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 $$\frac{dT}{dv}=2\mu v=2T/v$$

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