The period T of a simple pendulum with small oscillations is calculated from the formula, T =
2π/L/g, where L is the length of the pendulum and g is the acceleration due to gravity. Suppose that
values of L and g have errors of at most 0.2% and 0.05%, respectively. Use a linear approximation to approximate the maximum percentage error in the calculated value of T. Remember that the percentage errors can be high or low.

We are asked to approximate the maximum percentage error in the period \( T \) of a simple pendulum using a linear approximation. The period \( T \) is given by:

\[
T = 2\pi \sqrt{\frac{L}{g}}
\]

where:
- \( L \) is the length of the pendulum,
- \( g \) is the acceleration due to gravity.

The errors in \( L \) and \( g \) are at most 0.2% and 0.05%, respectively. We will use a linear approximation (differential method) to estimate the resulting error in \( T \).

### Step 1: Linear Approximation (Differential)
To use linear approximation, we first compute the total differential of \( T \). Taking the partial derivatives of \( T \) with respect to \( L \) and \( g \), we have:

\[
\frac{\partial T}{\partial L} = \frac{2\pi}{2} \cdot \frac{1}{\sqrt{gL}} = \frac{\pi}{\sqrt{gL}}
\]
\[
\frac{\partial T}{\partial g} = -\frac{2\pi}{2} \cdot \frac{L^{1/2}}{g^{3/2}} = -\frac{\pi L^{1/2}}{g^{3/2}}
\]

Now, the total differential \( dT \) is:

\[
dT = \frac{\partial T}{\partial L} dL + \frac{\partial T}{\partial g} dg
\]

Substituting the partial derivatives:

\[
dT = \frac{\pi}{\sqrt{gL}} dL - \frac{\pi L^{1/2}}{g^{3/2}} dg
\]

### Step 2: Relative Errors
We are given the percentage errors in \( L \) and \( g \). The relative errors are:

\[
\frac{dL}{L} = 0.002 \quad \text{(0.2% error in \( L \))}, \quad \frac{dg}{g} = 0.0005 \quad \text{(0.05% error in \( g \))}.
\]

Now, taking the relative error for \( T \):

\[
\frac{dT}{T} = \frac{1}{2} \frac{dL}{L} + \frac{1}{2} \frac{dg}{g}
\]

Substituting the given values:

\[
\frac{dT}{T} = \frac{1}{2} (0.002) + \frac{1}{2} (0.0005) = 0.001 + 0.00025 = 0.00125
\]

Thus, the maximum relative error in \( T \) is \( 0.00125 \), or 0.125%.

### Final Answer:
The maximum percentage error in the calculated value of \( T \) is **0.125%**.

---

Would you like further details or have any questions?

Here are 5 related questions to explore:
1. How does the period of a pendulum change with length \( L \)?
2. What is the effect of a higher gravitational constant \( g \) on the period \( T \)?
3. How would you calculate the error if the pendulum’s mass was also considered?
4. Can you derive the formula for a simple pendulum’s period from basic principles?
5. How would the error analysis change if the error in \( g \) were 1% instead of 0.05%?

**Tip:** In physics, percentage errors are always combined using a linear approximation when small errors are present, as it simplifies the error propagation process.

The period T of a simple pendulum with small oscillations is calculated from the formula, T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. Suppose that values of L and g have errors of at most 0.2% and 0.05%, respectively. Use a linear approximation to approximate the maximum percentage error in the calculated value of T.

Learn how to calculate the maximum percentage error in the period of a simple pendulum using linear approximation. This problem involves key concepts from calculus, including error propagation and differential approximation, with a detailed breakdown of formulas. Suitable for college-level students.

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