Let $g$ denote acceleration due to gravity. The period $\tau$ of a pendulum of length $r$ with small

oscillations is given by the formula

$$

\tau = 2\pi\sqrt{ \frac{r}{g} }.

$$

Suppose that experimental values of $r$ and $g$ have maximum errors of at most $0.5\%$ and $0.1\%$

respectively. Use differentials to approximate the maximum percentage error in the calculated value

of $\tau$. 

To approximate the maximum percentage error in the calculated value of the period $\tau$ of a pendulum using differentials, we can follow these steps:

### Step 1: Express $\tau$ in terms of $r$ and $g$

Given:

\[
\tau = 2\pi \sqrt{\frac{r}{g}}
\]

### Step 2: Compute the differential of $\tau$

The differential $d\tau$ approximates the change in $\tau$ due to small changes in $r$ and $g$. To find $d\tau$, we differentiate $\tau$ with respect to $r$ and $g$:

\[
d\tau = \frac{\partial \tau}{\partial r} dr + \frac{\partial \tau}{\partial g} dg
\]

First, find the partial derivatives:

\[
\frac{\partial \tau}{\partial r} = 2\pi \cdot \frac{1}{2} \cdot \left(\frac{r}{g}\right)^{-1/2} \cdot \frac{1}{g} = \frac{\pi}{\sqrt{rg}}
\]

\[
\frac{\partial \tau}{\partial g} = 2\pi \cdot \frac{1}{2} \cdot \left(\frac{r}{g}\right)^{-1/2} \cdot \left(-\frac{r}{g^2}\right) = -\frac{\pi r}{\sqrt{rg} \cdot g^2} = -\frac{\pi \sqrt{r}}{2 g^{3/2}}
\]

Thus:

\[
d\tau = \frac{\pi}{\sqrt{rg}} dr - \frac{\pi \sqrt{r}}{2 g^{3/2}} dg
\]

### Step 3: Relative Error Calculation

The relative error in $\tau$ is given by:

\[
\frac{d\tau}{\tau} = \frac{1}{\tau} \left( \frac{\partial \tau}{\partial r} dr + \frac{\partial \tau}{\partial g} dg \right)
\]

Substitute $\tau = 2\pi \sqrt{\frac{r}{g}}$ into the above expression:

\[
\frac{d\tau}{\tau} = \frac{1}{2\pi \sqrt{\frac{r}{g}}} \left( \frac{\pi}{\sqrt{rg}} dr - \frac{\pi \sqrt{r}}{2 g^{3/2}} dg \right)
\]

Simplifying:

\[
\frac{d\tau}{\tau} = \frac{1}{2} \cdot \frac{dr}{r} - \frac{1}{2} \cdot \frac{dg}{g}
\]

### Step 4: Maximum Percentage Error

Given that the maximum percentage errors in $r$ and $g$ are 0.5% and 0.1% respectively, we can substitute these values:

\[
\frac{dr}{r} = 0.5\% = 0.005
\]
\[
\frac{dg}{g} = 0.1\% = 0.001
\]

Substitute these into the relative error formula:

\[
\frac{d\tau}{\tau} = \frac{1}{2} \cdot 0.005 - \frac{1}{2} \cdot 0.001 = \frac{0.005}{2} - \frac{0.001}{2} = 0.002
\]

### Step 5: Express the Result as a Percentage

The maximum percentage error in $\tau$ is:

\[
\frac{d\tau}{\tau} \times 100\% = 0.002 \times 100\% = 0.2\%
\]

### Conclusion

The maximum percentage error in the calculated value of the period $\tau$ is approximately **0.2%**.

Would you like further details or have any questions? Here are some related questions:

1. How would the percentage error change if the error in $g$ were 0.5% instead of 0.1%?
2. What is the physical interpretation of the period $\tau$ in relation to the length $r$ and gravity $g$?
3. How does increasing the length $r$ of the pendulum affect the period $\tau$?
4. Can you derive the formula for the period $\tau$ using basic principles of physics?
5. How would the percentage error in $\tau$ be affected if $r$ and $g$ were measured with higher precision?

**Tip:** When using differentials to estimate errors, always ensure that the approximation is valid for small errors, as larger errors may require more sophisticated methods.

Learn how to approximate the maximum percentage error in the calculated period $ \tau $ of a pendulum using differentials. Understand the effects of small errors in length $ r $ and acceleration due to gravity $ g $ on the period formula. Suitable for advanced undergraduate students and researchers interested in error analysis in physics calculations.

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