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?

Question

Charles Anctil · Accepted Answer

#(dT)/T = 0.5%#

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%#

Cameron Alexander · Answer

#approx 1/2% # increase

see below

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

Charles Arocha · Answer

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

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

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.!

Paisley Johnson · Answer

undefined To find the increase in period when the length of a 1-foot pendulum is increased by 0.01 feet, you can use the formula for the period of a pendulum:

T = 3.14 * (L^0.5) / (2 * (2^0.5))

First, find the period of the original pendulum with a length of 1 foot:

T1 = 3.14 * (1^0.5) / (2 * (2^0.5))
   = 3.14 * (1) / (2 * 1.414)
   = 3.14 / 2.828
   ≈ 1.113 seconds

Now, find the period of the pendulum with a length increased by 0.01 feet:

New length = 1 + 0.01 = 1.01 feet

T2 = 3.14 * (1.01^0.5) / (2 * (2^0.5))
   ≈ 3.14 * 1.00498756211 / 2.828
   ≈ 1.114 seconds

The increase in period is the difference between T2 and T1:

Increase in period ≈ T2 - T1
                   ≈ 1.114 - 1.113
                   ≈ 0.001 seconds

So, the period increases by approximately 0.001 seconds.