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Let's analyze the problem step by step to determine the upper and lower bounds for the value of \( g \), the acceleration due to gravity.

We are given the formula for the time period \( T \) of a simple pendulum:

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

Where:
- \( T \) is the time period in seconds.
- \( l \) is the length of the pendulum in centimeters.
- \( g \) is the acceleration due to gravity in cm/s².
- \( \pi = 3.142 \).

Katie's experiment results are:
- \( l = 52.0 \, \text{cm} \) (correct to 3 significant figures).
- \( T = 1.45 \, \text{seconds} \) (correct to 3 significant figures).

### Rearranging the formula to find \( g \):
To solve for \( g \), rearrange the equation:

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

Square both sides:

\[
T^2 = 4\pi^2 \frac{l}{g}
\]

Now solve for \( g \):

\[
g = \frac{4\pi^2 l}{T^2}
\]

### Step 1: Calculate the bounds for \( l \) and \( T \)
Since \( l \) and \( T \) are given to 3 significant figures, the possible errors are:
- For \( l = 52.0 \, \text{cm} \), the error is \( \pm 0.05 \, \text{cm} \).
- For \( T = 1.45 \, \text{seconds} \), the error is \( \pm 0.005 \, \text{seconds} \).

So:
- The upper bound for \( l \) is \( 52.0 + 0.05 = 52.05 \, \text{cm} \).
- The lower bound for \( l \) is \( 52.0 - 0.05 = 51.95 \, \text{cm} \).

- The upper bound for \( T \) is \( 1.45 + 0.005 = 1.455 \, \text{seconds} \).
- The lower bound for \( T \) is \( 1.45 - 0.005 = 1.445 \, \text{seconds} \).

### Step 2: Calculate the upper and lower bounds for \( g \)

Now we calculate the value of \( g \) for both the upper and lower bounds.

#### Upper bound for \( g \):
\[
g_{\text{upper}} = \frac{4\pi^2 \times 52.05}{1.445^2}
\]

Substitute \( \pi = 3.142 \):

\[
g_{\text{upper}} = \frac{4 \times (3.142)^2 \times 52.05}{1.445^2}
\]

#### Lower bound for \( g \):
\[
g_{\text{lower}} = \frac{4\pi^2 \times 51.95}{1.455^2}
\]

Substitute \( \pi = 3.142 \):

\[
g_{\text{lower}} = \frac{4 \times (3.142)^2 \times 51.95}{1.455^2}
\]

I'll compute these values for you next.The calculated bounds for the acceleration due to gravity \( g \) are:

- Upper bound: \( g_{\text{upper}} = 984.37 \, \text{cm/s}^2 \)
- Lower bound: \( g_{\text{lower}} = 969.02 \, \text{cm/s}^2 \)

These values represent the range in which the value of \( g \) can be expected, based on the given measurements and their precision.

Would you like more details on any part of this calculation?

Here are some related questions to expand your understanding:
1. How does the precision of measurements affect the uncertainty in physical constants?
2. What is the relationship between the pendulum length and the period of oscillation?
3. How would using a different value of \( \pi \) affect the calculation?
4. How could experimental errors affect the final estimate of \( g \)?
5. Why is \( g \) expressed in cm/s² instead of m/s² in this context?

**Tip:** When working with uncertainties in experiments, always consider how small measurement errors can propagate through calculations to affect the final result.

Katie uses a simple pendulum in an experiment to estimate the value of g. Given l = 52.0 cm and T = 1.45 seconds (both correct to 3 significant figures), calculate the upper and lower bounds for g using T = 2π√(l/g). Use π = 3.142.

Learn how to calculate the upper and lower bounds of the gravitational acceleration (g) using the period of a pendulum. The problem involves determining the bounds based on experimental values of l = 52.0 cm and T = 1.45 s, with detailed steps and formulas.

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