need help please. the time required for one complete oscillation of a pendulum is called its period. if L is the length of the pendulum, then the period is given by P= 2 times pie multiplied by the square root of (L/g), where g is a constant called the acceleration due to gravity. use differentials to show that the percentage error in P is approximately half the percentage error in L.
P=2πsqrt(L/g)
dP/dL = 2π(1/2)L^(-1/2)/sqrt(g)
=2πsqrt(L/g)/(2L)
transpose appropriate terms:
2*[dP/(2πsqrt(L/g))] = [dL/(L)]
Each term in square brackets is equal to the fractional form of percentage error, thus proved.
(use Δ => differentials instead of derivative.)
To find the percentage error in P and show that it is approximately half the percentage error in L, we can use differentials. Let's go through the steps:
1. First, let's find the differential of P with respect to L.
- The period is given by the equation P = 2π√(L/g).
- Taking the differential of both sides, we get dP = d(2π√(L/g)).
- Using the chain rule, the differential becomes dP = (2π/2√(L/g)) * (1/2) * g^-1/2 * dL.
- Simplifying, dP = π * g^-1/2 * dL.
2. Next, let's find the percentage error in P.
- The percentage error in P, denoted as ΔP, is given by ΔP/P * 100%.
- Let's assume there is a small change in the length L, denoted as ΔL.
- The percentage error in L, denoted as ΔL, is given by ΔL/L * 100%.
- We can write ΔL/L as dL/L, where dL is the differential change in L.
3. Now, let's substitute the differential of P (dP) and the differential change in L (dL) into the percentage error equations.
- ΔP/P * 100% = (dP/P) * 100% = (π * g^-1/2 * dL / (2π√(L/g))) * 100%.
- Simplifying, ΔP/P * 100% = g^-1/2 * (dL / √(L/g)) * 100%.
4. Finally, let's simplify the equation and make the comparison between the percentage errors in P and L.
- ΔP/P * 100% = g^-1/2 * (dL / √(L/g)) * 100%.
- ΔP/P * 100% = g^-1/2 * √(g/L) * (dL / L) * 100%.
- Notice that g^-1/2 * √(g/L) is a constant.
- Let's denote this constant as k. So, the equation becomes ΔP/P * 100% = k * (dL / L) * 100%.
- Therefore, the percentage error in P is proportional to the percentage error in L, with k as the proportionality constant.
In conclusion, we have shown that the percentage error in P is approximately half the percentage error in L, as they are proportional with a constant of proportionality.