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Chapter 13: Problem 12

In the measurcment of \(g\) using simple pendulum the time taken for 30 oscillations is \((61.2 \pm 1.3) \mathrm{s}\) and cffective length of pendulum is \((1.00 \pm 0.01) \mathrm{m} .\) The value of \(g\) with error limits is: (a) \((9.80 \pm 0.2) \mathrm{m} / \mathrm{s}^{2}\) (b) \((9.81 \pm 0.47) \mathrm{m} / \mathrm{s}^{2}\) (c) \((9.49 \pm 0.1) \mathrm{m} / \mathrm{s}^{2}\) (d) \((9.49 \pm 0.47) \mathrm{m} / \mathrm{s}^{2}\)

Short Answer

Expert verified
The value of \( g \) is \((9.49 \pm 0.47) \ \text{m/s}^2\), matching option (d).

Step by step solution

01

Calculate Time Period

The time period \( T \) for one oscillation is calculated using \( T = \frac{t}{n} \), where \( t = 61.2 \) seconds is the total time for \( n = 30 \) oscillations. Thus, \( T = \frac{61.2}{30} = 2.04 \, \text{s} \).
02

Determine the Uncertainty in Time Period

Calculate the uncertainty in \( T \) as \( \Delta T = \frac{\Delta t}{n} = \frac{1.3}{30} = 0.0433 \, \text{s} \). So, \( T = 2.04 \pm 0.0433 \text{s} \).
03

Use the Formula for Acceleration Due to Gravity

The formula for \( g \) using a simple pendulum is \( g = \frac{4\pi^2 L}{T^2} \). Here, \( L = 1.00 \pm 0.01 \) meters and \( T = 2.04 \pm 0.0433 \) seconds.
04

Calculate \( g \)

Use the formula to calculate \( g \): \[ g = \frac{4\pi^2 \times 1.00}{(2.04)^2} = 9.49 \ \text{m/s}^2 \].
05

Determine the Uncertainty in \( g \)

The relative error in \( g \) is given by \( \frac{\Delta g}{g} = \sqrt{\left( \frac{\Delta L}{L} \right)^2 + \left( 2 \times \frac{\Delta T}{T} \right)^2} \). Substituting the given values, we find: \[ \frac{\Delta g}{g} = \sqrt{\left( \frac{0.01}{1.00} \right)^2 + \left( 2 \times \frac{0.0433}{2.04} \right)^2} = 0.0495 \].
06

Calculate the Final Error in \( g \)

Thus, \( \Delta g = 9.49 \times 0.0495 = 0.47 \). Therefore, \( g = 9.49 \pm 0.47 \ \text{m/s}^2 \).
07

Identify the Correct Option

The calculated value with error is \( (9.49 \pm 0.47) \ \text{m/s}^2 \), matching option (d).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oscillation
In the context of a simple pendulum, oscillation refers to the repetitive motion that the pendulum undergoes about its equilibrium position. When released from a height, the pendulum swings back and forth under the influence of gravity.
This continual to-and-fro motion is surprisingly consistent and can be described by physics.
  • Amplitude: The maximum extent of the swing from the central (equilibrium) position.
  • Frequency: How often the pendulum completes an oscillation in a given time, typically measured in Hertz (Hz).
  • Time Period: The time it takes for one complete oscillation.
Understanding oscillation is key to grasping how simple pendulums are used for measuring gravitational acceleration. Their predictable movement allows us to derive important formulas, like those for the time period and ultimately, the calculation of gravitational acceleration.
Gravitational Acceleration Measurement
The measurement of gravitational acceleration, denoted as "g", using a simple pendulum relies on the pendulum's time period and length. The formula used is:
\[ g = \frac{4\pi^2 L}{T^2} \]
Here, L is the length of the pendulum, and T is the time period for one oscillation. By measuring how long it takes the pendulum to complete a set number of oscillations, we can calculate T and subsequently determine g.
  • The formula comes from mathematical physics, where the combination of constants \( 4\pi^2 \) ensures appropriate unit conversion.
  • This method provides a relatively simple approach to measuring gravitational acceleration due to its reliance on basic observations.
The accuracy of this measurement ensures that many scientific investigations can account for local variations in gravitational acceleration, important for fields like geophysics and metrology.
Error Propagation
Error propagation is an important concept in experimental physics and mathematics. It helps us understand how measurement uncertainties affect calculated results. In the context of the simple pendulum experiment, any small errors in measuring time or length can propagate and affect the final result of gravitational acceleration.
  • The formula for relative error in g is: \[ \frac{\Delta g}{g} = \sqrt{\left( \frac{\Delta L}{L} \right)^2 + \left( 2 \times \frac{\Delta T}{T} \right)^2} \]
  • This formula comprehensively evaluates how uncertainties in both L and T combine to affect the accuracy of g.
  • It's crucial to use accurate measurements and minimize sources of error to improve the reliability of the calculated gravitational acceleration.
Understanding error propagation is crucial for improving experimental accuracy and ensuring that scientific results are valid and reliable.
Time Period Calculation
Calculating the time period of a pendulum is a straightforward yet essential step in using pendulums for practical measurements. The time period \( T \), or the time it takes to complete one full oscillation, is determined by dividing the total time taken for multiple oscillations by the number of oscillations:
\[ T = \frac{t}{n} \]
Here, \( t \) is the total time, and \( n \) is the number of oscillations. In our exercise, 30 oscillations took 61.2 seconds, resulting in a time period of 2.04 seconds.
  • Knowing the time period allows physicists to relate the pendulum's physical properties to gravitational acceleration.
  • Even small errors in timing can influence the outcome, highlighting the need for precise timekeeping.
Grasping this fundamental concept ensures that students can effectively use pendulums for various applications, such as determining local gravitational constants.

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Most popular questions from this chapter

In an cxperiment to determinc the cocfficient of viscosity of a liquid using the formula \(\eta-\frac{\pi \operatorname{Pr}^{4}}{8 l Q}\) the radius of the capillary tube was measured to be \(0.41 \mathrm{~mm}\) using an instrument of least count \(0.001 \mathrm{~cm}\), length \(l\) was measured using a metre scale as \(15 \mathrm{~cm}\). The percentage error in the pressure difference is \(0.4 \%\) and the percentage error in the volume of the liquid flowing out per second \(Q\) is \(0.3 \%\) what is the maximum percentage error in the coefficient of viscosity?

\(\Lambda\) physical quantity \(X\) is related to three observable \(a, b\) and \(c\) as \(X-\frac{b^{2} \sqrt{a}}{c} .\) The errors of measurements in \(a, b\) and \(c\) are \(4 \%, 3 \%\) and \(2 \%\) respoctively. What is the percentage crror in the quantity \(X^{n}\) ?

A parlicle is moving on a straight linc with a speed of about \(0.5 \mathrm{~m} / \mathrm{s}\). To measure its speed, points are marked on the linc of motion, which are cxactly separated by \(5 \mathrm{~mm}\). To have an accuracy of \(1 \%\) in the dctermination of specd, the minimum least count of the measuring clock is (a) \(0.1 \mathrm{~s}\) (b) \(0.01 \mathrm{~s}\) (c) \(0.001 \mathrm{~s}\) (d) \(0.000 \mathrm{I} \mathrm{s}\)

If forec \(F\), acceleration \(A\) and time \(T\) be taken as the fundamental physical quantities, the dimensions of length on this system of units are (a) \(F A T^{2}\) (b) \(F A T\) (c) \(F T\) (d) \(A T^{2}\)

Solve the following with regard to the correct significant figures (a) \(\sqrt{4.8-2.35}\) (b) \(\left(2.35 \times 10^{3}-1.2 \times 10^{3}\right) / 2.0 \times 10^{2}\)
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