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.
To calculate the percentage error in the period (P) of a pendulum, we need to use differentials. Let's start by finding the differential expression for the period.
Given: P = 2π(L/g)^(1/2)
Taking the natural logarithm (ln) of both sides, we get:
ln(P) = ln(2π(L/g)^(1/2))
Now, we can differentiate both sides of the equation with respect to L. However, before doing that, let's simplify the equation by applying logarithmic properties:
ln(P) = ln(2π) + 1/2 ln(L/g)
Now, differentiating both sides with respect to L, we have:
d(ln(P))/dL = 1/2d(ln(L/g))/dL
Since d(ln(x))/dx = 1/x, we can substitute:
d(ln(P))/dL = (1/2) * (1/L)
Since d(ln(P))/dL is the relative change in the period (P) with respect to a relative change in L, we can express it as a percentage:
(d(ln(P))/dL) * (L/P) * 100 = (1/2L) * (L/P) * 100
= 1/2 * (1/P) * 100
= 1/2 * (100/P)
Now, the relative change in the period is the same as the percentage error in P:
Percentage error in P = (1/2) * (100/P)
Next, we need to calculate the percentage error in L. Let's assume ΔP and ΔL are the absolute errors in P and L, respectively. Therefore, we can write:
Percentage error in L = (ΔL/L) * 100
Now, we can establish a relationship between the percentage errors in P and L:
(Percentage error in P) / (Percentage error in L) = [(1/2) * (100/P)] / [(ΔL/L) * 100]
= (1/2) * (1/P) * (L/ΔL)
= (1/2) * [ (L/ΔL) / P ]
≈ (1/2) * (dL/dP)
This shows that the percentage error in P is approximately half the percentage error in L when differentials are used.
Please note that this approximation holds when the fractional errors in L and P are small, and it assumes that the magnitude of the errors in L and P are comparable.
To show that the percentage error in P is approximately half the percentage error in L, we need to find the differentials of P and L and compare their magnitudes.
Let's start by finding the differential of P. We can use the chain rule for differentiation:
dP = ∂P/∂L * dL,
where ∂P/∂L represents the partial derivative of P with respect to L and dL represents the differential of L.
Taking the derivative of the given equation for P with respect to L:
∂P/∂L = 2π * (1/2) * (L/g)^(-1/2) * (1/g),
Simplifying this expression, we get:
∂P/∂L = π / (gL)^(1/2).
Now, let's find the differential of L:
dL = ∂L/∂L * dL = dL,
Since ∂L/∂L equals 1.
To compare the magnitudes of dP and dL, we'll take the absolute value of both.
|dP| = |π / (gL)^(1/2) * dL| = |π / (gL)^(1/2)| * |dL|,
Assuming the percentage error in L is given by ε₁, we have:
ε₁ = (|dL| / L) * 100%.
Substituting this expression into our equation for |dP|, we get:
|dP| = |π / (gL)^(1/2)| * ε₁.
To find the percentage error in P, we divide |dP| by P and multiply by 100%:
ε₂ = (|dP| / P) * 100%.
Substituting the equation for |dP| and P in terms of L, we have:
P = 2π(L/g)^(1/2),
ε₂ = (|π / (gL)^(1/2)| * ε₁) / (2π(L/g)^(1/2)) * 100%,
Simplifying this expression, we get:
ε₂ = (ε₁ / 2) * 100%.
Therefore, the percentage error in P (represented by ε₂) is approximately half of the percentage error in L (represented by ε₁).