P = 2/√g √L
dP = 2/√g * 1/2√L dL
= 1/√(gL) dL
So, the % error in P is
dP/P = 1/√(gL) dL/P
= 1/√(gL) * √g/2√L dL
= 1/√g * √g/2 dL/L
= 1/2 dL/L
the time required for one complete oscillation of a pendulum is called its period. if the length L of the pendulum is measured in feet and the period P n seconds, then the period is given by P=2¶(L/g)^1/2, where g is constant. use differentials to show the percentage error in P is approximately half the percentage error in L.
1 answer
1 answer