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?

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

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.!

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%#