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A simple pendulum of length \(1.53 \mathrm{~m}\) makes \(72.0\) oscillations in \(180 \mathrm{~s}\) at a certain location. Find the acceleration due to gravity at this point.

Short Answer

Expert verified
The acceleration due to gravity at this point is approximately \( 9.81 \mathrm{~m/s^2} \).

Step by step solution

01

Understand the problem and identify given parameters

We are given the length \( L \) of the pendulum as 1.53m, and we know that it makes 72.0 oscillations in 180s. We are tasked with finding the acceleration due to gravity \( g \).
02

Calculate the period of the pendulum

The period of the pendulum (T) is the time it takes for one full oscillation. We can calculate this by dividing the total time by the number of oscillations. That is, \( T = \frac{180s}{72.0} = 2.5s \).
03

Use the period formula to find the acceleration due to gravity

Pendulum motion is governed by the formula \( T = 2\pi\sqrt{\frac{L}{g}} \). We can rearrange this formula to find \( g \), the acceleration due to gravity: \( g = \frac{4\pi^2L}{T^2}\).
04

Substituting the known values into the formula

Substitute 1.53 for \( L \), and 2.5 for \( T \) into the formula \( g = \frac{4\pi^2L}{T^2}\), yielding that \( g = \frac{4\pi^2 \times 1.53}{2.5^2}\).
05

Calculate the acceleration due to gravity

Solving the formula results in \( g \approx 9.81 \mathrm{~m/s^2} \)

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Key Concepts

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

Simple Pendulum
A simple pendulum consists of a weight, often referred to as the 'bob,' suspended by a string of fixed length, which allows it to swing back and forth in an arc. This arrangement is fundamental in physics, as it exhibits periodic motion, where the bob moves from its highest point (the amplitude of the swing) down through the equilibrium point and to the opposite side symmetrically. It's this to-and-fro motion that creates oscillations.

Why is the simple pendulum significant? The beauty of the simple pendulum lies in its predictability and simplicity. It serves as a classic example of harmonic motion - the type of motion we see when the restoring force is directly proportional to the displacement from an equilibrium position. The pendulum’s movement is also influenced by the acceleration due to gravity, which makes it an invaluable tool in measuring this fundamental force of nature, as demonstrated in the exercise provided.
Oscillations
Essentially, oscillations refer to the repeated back and forth motion of the pendulum bob over a central point, or its equilibrium position. In a single cycle, the bob swings from one side, reaches the maximum height on the opposite side (making one complete oscillation), and then returns. The pendulum keeps doing this at a consistent frequency, assuming no external forces such as air resistance affect it significantly.

The rate of oscillation is a critical aspect of pendulum motion and is determined by factors like the length of the pendulum and the local gravitational field strength. This precise, rhythmic motion can be utilized in clock mechanisms for timekeeping, and in the sciences, to help calculate gravitational forces. As in our exercise, the number of oscillations observed provides key data for calculating further properties of the pendulum's motion.
Period of Pendulum
The period of a pendulum is the time it takes to complete one full oscillation. It's a crucial parameter that is relatively simple to measure and can reveal much about the conditions under which the pendulum is swinging. The period is affected by the length of the pendulum and the acceleration due to gravity at its location, thereby providing a pathway to measure the elusive gravitational constant on Earth.

Using the relationship given by the formula \( T = 2\pi\sqrt{\frac{L}{g}} \), where \( T \) is the period, \( L \) is the length of the pendulum, and \( g \) is the acceleration due to gravity, we can manipulate the equation to find any one of these values if the other two are known. It's important to remember that factors like air resistance and the amplitude of swing are considered negligible in the idealized mathematical model of a simple pendulum. If precision is imperative, these factors can affect the period and should be accounted for in a more detailed analysis, not unlike how we improved our understanding of the period to find the acceleration due to gravity in our exercise.

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Most popular questions from this chapter

An automobile can be considered to be mounted on four springs as far as vertical oscillations are concerned. The springs of a certain car of mass \(1460 \mathrm{~kg}\) are adjusted so that the vibrations have a frequency of \(2.95 \mathrm{~Hz}\). (a) Find the force constant of each of the four springs (assumed identical). (b) What will be the vibration frequency if five persons, averaging \(73.2\) kg each, ride in the car?

An engineer wants to find the rotational inertia of an oddshaped object of mass \(11.3 \mathrm{~kg}\) about an axis through its center of mass. The object is supported with a wire through its center of mass and along the desired axis. The wire has a torsional constant \(\kappa=0.513 \mathrm{~N} \cdot \mathrm{m} .\) The engineer observes that this torsional pendulum oscillates through \(20.0\) cycles in \(48.7 \mathrm{~s}\). What value of the rotational inertia is calculated?

An oscillating block-spring system has a mechanical energy of \(1.18 \mathrm{~J}\), an amplitude of \(9.84 \mathrm{~cm}\), and a maximum speed of \(1.22 \mathrm{~m} / \mathrm{s} .\) Find \((a)\) the force constant of the spring, \((b)\) the mass of the block, and \((c)\) the frequency of oscillation.

A U-tube is filled with a single homogeneous liquid. The liquid is temporarily depressed in one side by a piston. The piston is removed and the level of liquid in each side oscillates. Show that the period of oscillation is \(\pi \sqrt{2 L / g}\), where \(L\) is the total length of the liquid in the tube.

Sketch the path of a particle that moves in the \(x y\) plane according to \(x=x_{\mathrm{m}} \cos (\omega t-\pi / 2)\) and \(y=2 x_{\mathrm{m}} \cos \omega t\)

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