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.

Question

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.

Answer 1

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

Answer 2

To find the percentage error in P, we need to first find the differential of P with respect to L. Let's differentiate the equation P = 2π(L/g)^(1/2) with respect to L.

dP/dL = d/dL [2π(L/g)^(1/2)]
= 2π * (1/2) * (L/g)^(-1/2) * (1/g) [By applying the power rule of differentiation]
= π * (L/g)^(-1/2) * (1/g)
= π/g * (L/g)^(-1/2)

The percentage error in P, denoted as δP/P, can be approximated using the concept of differentials. We divide both sides of the equation by P:

δP/P = (π/g * (L/g)^(-1/2)) * (δL/L)

To find the percentage error in L, δL/L, we can rearrange the given equation:

P = 2π(L/g)^(1/2)

Raising both sides to the power of 2:

P^2 = 4π^2(L/g)

Differentiating both sides with respect to L gives us:

2P * dP/dL = 4π^2/g

Simplifying:

dP/dL = 2π^2/gP

Now, we can solve for δL/L by dividing both sides by dL:

(δL/L) = (dP/dL) / (2π^2/gP)
= (2π/g * (L/g)^(-1/2)) / (2π^2/gP)
= (L/g)^(-1/2) / (πP)

Substituting this expression for δL/L back into the equation for δP/P, we get:

δP/P = (π/g * (L/g)^(-1/2)) * (L/g)^(-1/2) / (πP)
= (L/g)^(-1) * (L/g)^(-1/2) / P
= (L/g)^(-3/2) / P

The absolute value of the percentage error in P can be expressed as |δP/P|, so the final equation becomes:

|δP/P| = (L/g)^(-3/2) / P

Now, let's analyze the expression:

|δP/P| equals (L/g)^(-3/2) divided by P. Notice that (L/g)^(-3/2) is the reciprocal of (L/g)^(3/2). Since P is the period, it is always positive. Thus, the percentage error in P is always positive.

We can observe that (L/g)^(-3/2) is inversely proportional to (L/g)^(3/2). Suppose we have the percentage error in L, which we can denote as δL/L. Then, the percentage error in P is approximately equal to half of δL/L. Mathematically, we can represent this as:

|δP/P| ≈ 0.5 * |δL/L|

Therefore, we can conclude that the percentage error in P is approximately half the percentage error in L.