Math Problem Statement

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.

Solution

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

We are given the formula for the time period TT of a simple pendulum:

T=2πlgT = 2\pi \sqrt{\frac{l}{g}}

Where:

  • TT is the time period in seconds.
  • ll is the length of the pendulum in centimeters.
  • gg is the acceleration due to gravity in cm/s².
  • π=3.142\pi = 3.142.

Katie's experiment results are:

  • l=52.0cml = 52.0 \, \text{cm} (correct to 3 significant figures).
  • T=1.45secondsT = 1.45 \, \text{seconds} (correct to 3 significant figures).

Rearranging the formula to find gg:

To solve for gg, rearrange the equation:

T=2πlgT = 2\pi \sqrt{\frac{l}{g}}

Square both sides:

T2=4π2lgT^2 = 4\pi^2 \frac{l}{g}

Now solve for gg:

g=4π2lT2g = \frac{4\pi^2 l}{T^2}

Step 1: Calculate the bounds for ll and TT

Since ll and TT are given to 3 significant figures, the possible errors are:

  • For l=52.0cml = 52.0 \, \text{cm}, the error is ±0.05cm\pm 0.05 \, \text{cm}.
  • For T=1.45secondsT = 1.45 \, \text{seconds}, the error is ±0.005seconds\pm 0.005 \, \text{seconds}.

So:

  • The upper bound for ll is 52.0+0.05=52.05cm52.0 + 0.05 = 52.05 \, \text{cm}.

  • The lower bound for ll is 52.00.05=51.95cm52.0 - 0.05 = 51.95 \, \text{cm}.

  • The upper bound for TT is 1.45+0.005=1.455seconds1.45 + 0.005 = 1.455 \, \text{seconds}.

  • The lower bound for TT is 1.450.005=1.445seconds1.45 - 0.005 = 1.445 \, \text{seconds}.

Step 2: Calculate the upper and lower bounds for gg

Now we calculate the value of gg for both the upper and lower bounds.

Upper bound for gg:

gupper=4π2×52.051.4452g_{\text{upper}} = \frac{4\pi^2 \times 52.05}{1.445^2}

Substitute π=3.142\pi = 3.142:

gupper=4×(3.142)2×52.051.4452g_{\text{upper}} = \frac{4 \times (3.142)^2 \times 52.05}{1.445^2}

Lower bound for gg:

glower=4π2×51.951.4552g_{\text{lower}} = \frac{4\pi^2 \times 51.95}{1.455^2}

Substitute π=3.142\pi = 3.142:

glower=4×(3.142)2×51.951.4552g_{\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 gg are:

  • Upper bound: gupper=984.37cm/s2g_{\text{upper}} = 984.37 \, \text{cm/s}^2
  • Lower bound: glower=969.02cm/s2g_{\text{lower}} = 969.02 \, \text{cm/s}^2

These values represent the range in which the value of gg 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 π\pi affect the calculation?
  4. How could experimental errors affect the final estimate of gg?
  5. Why is gg 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.

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Math Problem Analysis

Mathematical Concepts

Physics
Gravitational Acceleration
Significant Figures
Upper and Lower Bounds

Formulas

T = 2π√(l/g)
g = (4π²l) / T²

Theorems

Error Propagation
Uncertainty in Measurements

Suitable Grade Level

Grades 11-12