Question

A pendulum of length L feet has a period of: `T= 3.14(L^.5)/2(2^.5)` seconds. If you increase the length of a 1-foot pendulum by .01 feet, how much does its period increase?

Answer 1

`approx 1/2%` increase

see below

Explanation:

I'd normally expect `T = 2pi sqrt(L/g)`

you seem to have

`T(L) = pi/2 sqrt(2L)`, which is fine, might be to do with old imperial units and maybe a different gravity. we can work with that

So with `T_o = pi/2 sqrt(2L)` you want to know

`T(L + delta L) = pi/2 sqrt(2(L+ delta L))`

but by approximation

We can plough the same furrow as Newton (and Taylor expansions), but in a way that is simpler, using a generalised binomial expansion, ie this idea

`(1+ h)^alpha = 1 + alpha h + (alpha(alpha-1))/(2!) h^2 + O(h^3) qquad abs h < 1 qquad alpha in mathcal(R), notin mathcal(Z^+)`

It looks just like a Taylor Expansion, ie based in Calculus. In fact I think it was Newton that generalised the binomial expansion in this way.

So

`T(L + delta L) = pi/2 sqrt(2(L+ delta L))`

`= pi/2 sqrt(2L) (1+ (delta L)/L)^(1/2) " where "qquad abs ((delta L)/L) " << " 1`

which is simply

`T_o (1+ (delta L)/L)^(1/2)`

`=T_o (1+ 1/2 (delta L)/L + O(deltaL^2) )`

`approxT_o (1+ 1/2 (delta L)/L )`

`= T_o (1+ (delta L)/(2L) )`

`= T_o (1+ 0.005 )`

That is a c.1/2% increase

Answer 2

App. increase in period `T` is `(pisqrt2)/400 "sec"`.

`"App." % "increase in" T=1/2 %`.

Explanation:

The Period `T` (in sec.) of a pendulum of Length `L` (in feet) is given by the Formula,

`T=(pi/2)sqrt(2L)`

Let `deltaT` denote increase in `T`, and, `delta L` in `L`,

Then, from Calculus, we know that, `deltaT~=(dT)/(dL)*deltaL...(star)`

`T=(pi/2)sqrt(2L) rArr (dT)/(dL)=((pisqrt2)/2)(1/(2sqrtL))=(pisqrt2)/(4sqrtL)`.

`:. deltaT~=(pisqrt2)/(4sqrtL)deltaL.............[by(star)]`

Here, `L=1, deltaL=0.01`

`rArr deltaT~=(pisqrt2)/4*0.01=(pisqrt2)/400 "feet"`.

Thus, app. increase in period `T` is `(pisqrt2)/400 "sec"`.

Note that, when `L=1 "foot, period (initial)" T=(pisqrt2)/2 sec.`

Hence, `"app." % "increase in" T=((deltaT)/T)100`

`=((pisqrt2)/400)/((pisqrt2)/2)*100=1/2%`, as Respected Eddie has obtained !

Enjoy Mats.!

Answer 3

`(dT)/T = 0.5%`

Explanation:

Error propagation is easily handled using the `log` transformation.
Given `T=pi/2sqrt(L/g)` after tansformation reads

`log_eT = log_e(pi/2) +1/2(log_eL-log_e g)`

After deriving we have

`(dT)/T = 1/2((dL)/L-(dg)/g)`

supposing that `g` is known with precision `(dg)/g approx 0`

so

`(dT)/T = 1/2(dL)/L = 1/2(0.01)/1 = 0.005`

or also

`(dT)/T = 0.5%`