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?

3 Answers
Aug 26, 2016

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

Aug 26, 2016

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.!

Aug 26, 2016

(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%