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A small object oscillates back and forth at the bottom of a frictionless hemispherical bowl, as the drawing illustrates. The radius of the bowl is \(R\), and the angle \(\theta\) is small enough that the object oscillates in simple harmonic motion. Derive an for the angular frequency \(\omega\) of the motion. Express your answer in terms of \(R\) and \(g,\) the magnitude of the acceleration due to gravity.

Short Answer

Expert verified
The angular frequency is \( \omega = \sqrt{\frac{g}{R}} \).

Step by step solution

01

Understand the motion

The object is oscillating at the bottom of a frictionless hemispherical bowl, and this motion can be approximated as simple harmonic motion (SHM) for small angles \( \theta \). Simple harmonic motion can typically be described by oscillations where a restoring force is proportional to displacement.
02

Identify forces involved

For small angles \( \theta \), the restoring force is due to the gravitational component acting along the surface of the bowl. This force can be expressed as \( mg\sin\theta \). For SHM, we use the small angle approximation \( \sin\theta \approx \theta \), hence the restoring force becomes \( mg\theta \).
03

Express displacement geometrically

Consider the arc as the path of displacement, \( s = R\theta \). The displacement from equilibrium (down the bowl) is proportional to \( R\theta \). The force of gravity along the arc is simplifying to \( F = -mg\theta \approx -mgs/R \).
04

Apply Newton's Second Law

Using Newton's second law along the direction of motion, we have \( ma = -mgs/R \). Since \( a \) is the linear acceleration and can be related to the angular frequency \( a = \omega^2 s \), substitute to get \( m\omega^2 s = mgs/R \).
05

Solve for angular frequency \( \omega \)

Simplify the equation \( m\omega^2 R\theta = mg\theta \) to derive \( \omega^2 = g/R \) and thus \( \omega = \sqrt{g/R} \).
06

Final expression for \( \omega \)

The angular frequency \( \omega \) of the simple harmonic motion is given by \( \omega = \sqrt{\frac{g}{R}} \). This completes the derivation.

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Key Concepts

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

Angular Frequency
Angular frequency, often represented by the Greek letter \( \omega \), is a key concept in understanding simple harmonic motion (SHM). It describes how quickly an object oscillates through its cycle. Angular frequency is akin to the regular frequency, but instead of cycles per unit time, it refers to radians per unit time. This can be expressed mathematically as the number of complete oscillations in radians the object moves through per second.

In the exercise we looked at, angular frequency \( \omega \) is derived from the object's movement in a frictionless hemispherical bowl. It oscillates back and forth following the rules of SHM. For small displacements from the center, the formula for angular frequency simplifies to:

\[ \omega = \sqrt{\frac{g}{R}} \]

where \( g \) is the gravitational acceleration (approximately \( 9.81 \, \text{m/s}^2 \) on Earth's surface) and \( R \) is the radius of the bowl.

This relationship shows that the angular frequency solely depends on the gravitational force and the bowl's radius. Decreasing the radius or changing the gravitational acceleration will directly affect how quickly the object oscillates.
Newton's Second Law
Newton's Second Law is pivotal for understanding motion dynamics. It states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass. Mathematically, it is represented as:

\[ F = ma \]

where \( F \) is the force applied, \( m \) is the mass of the object, and \( a \) is the acceleration.

In the scenario with the hemispherical bowl, this law is essential in deriving the angular frequency of oscillation. For small angles \( \theta \), the object's restoring force can be considered through the gravitational force component acting sideways. Using Newton's Second Law, this force is expressed as:

\[ ma = -mg\frac{s}{R} \]

We substitute the linear force relationship \( F = -mg\theta \) with \( s = R\theta \) to relate these variables, rendering the equation as:

\[ m\omega^2 s = mg\frac{s}{R} \]

This substitution simplifies the arrangement and provides a pathway to derive \( \omega \). Newton’s law forms the backbone of understanding how forces translate to motions in systems like these.
Restoring Force
Restoring force is a fundamental concept of simple harmonic motion, helping to maintain the oscillation by pulling the system back towards equilibrium. In the context of our scenario, the restoring force is what pulls the object back towards the lowest point of the hemispherical bowl.

Here, the key indicator for restoring force is gravity acting at an angle. For small oscillations, we use the approximation \( \sin\theta \approx \theta \), allowing us to simplify the restoring force as:

\[ F = mg\theta \]

In the exercise, the arc length \( s \) acts as the path of the object's displacement, meaning that \( s = R\theta \), and thus the restoring force becomes proportional to this arc displacement. The force becomes a linear approximation:

\[ F = -\frac{mg}{R} s \]

It's this force that balances the gravitational pull, tethering the object to its motion path while allowing it to oscillate continuously. Thus, the equation illustrates how restoring force is related to both gravity's influence and the geometrical parameters of the system, like radius \( R \).

Understanding how restoring force works is critical, as it keeps the system perpetually adjusting back towards a state of equilibrium, characteristic of oscillatory systems in SHM.

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Most popular questions from this chapter

A 1.1-kg object is suspended from a vertical spring whose spring constant is \(120 \mathrm{N} / \mathrm{m}\). (a) Find the amount by which the spring is stretched from its unstrained length. (b) The object is pulled straight down by an additional distance of \(0.20 \mathrm{m}\) and released from rest. Find the speed with which the object passes through its original position on the way up.

Two physical pendulums (not simple pendulums) are made from meter sticks that are suspended from the ceiling at one end. The sticks are uniform and are identical in all respects, except that one is made of wood (mass \(=0.17 \mathrm{kg}\) ) and the other of metal (mass \(=0.85 \mathrm{kg}\) ). They are set into oscillation and execute simple harmonic motion. Determine the period of (a) the wood pendulum and (b) the metal pendulum.

In 0.750 s, a 7.00-kg block is pulled through a distance of 4.00 m on a frictionless horizontal surface, starting from rest. The block has a constant acceleration and is pulled by means of a horizontal spring that is attached to the block. The spring constant of the spring is 415 N/m. By how much does the spring stretch?

To measure the static friction coefficient between a 1.6-kg block and a vertical wall, the setup shown in the drawing is used. A spring (spring constant = 510 N/m) is attached to the block. Someone pushes on the end of the spring in a direction perpendicular to the wall until the block does not slip downward. The spring is compressed by 0.039 m. What is the coefficient of static friction?

A 0.70-kg block is hung from and stretches a spring that is attached to the ceiling. A second block is attached to the fi rst one, and the amount that the spring stretches from its unstrained length triples. What is the mass of the second block?

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