# Oscillations of beat in tube

• Numeriprimi
In summary, the problem involves a small bead with mass m and charge q, free to move in a horizontal tube without friction. The tube is placed between two spheres with opposite charges Q = -q, separated by a distance of 2a. The frequency of small oscillations around the equilibrium point of the bead is being sought. Using Coulomb's law, the net force on the bead at an arbitrary x can be determined, and from there the effect of the force and the frequency of oscillations can be calculated.

## Homework Statement

Bead (mass point - mass m and charge q) is free to move in a horizontal tube (without friction). The tube is between two spheres (separated by a distance 2a) with charges Q = − q. What is the frequency of small oscillations around the equilibrium point of the bead?

Coulomb's law

## The Attempt at a Solution

Hmm... Coulomb's law defines electrostatic interaction between two spheres. The spheres have same charges - repel, but the bead have opposite charge, so the spheres and repel attract. It causes oscillation? Then we may use an equation for the mechanical oscillation... I don't know how combine mechanical oscillation and electrostatic interaction in one equation.
Can you help me?

Thank you very much and sorry for my bad English.

Obviously, midway between the spheres the forces from both spheres cancel each other out, so at this point the bead is in equilibrium. Let x be the coordinate of the bead, and let x = 0 at the equilibrium. What is the net force on the bead at an arbitrary x?

Are you sure that the charges at the ends of the tube have sign opposite to the charge in the middle?

Can you put the actual text of the problem? Or diagram.
The way I understand it, there will be no oscillations.
The middle point is indeed an equilibrium position but when the bead is moved from equilibrium the net force will be directed away from equilibrium.

Ok, I may write it bad, so I'm going to write it better and I take you a picture...

A small bead of mass m and charge q is free to move in a horizontal tube. The tube is placed in between two spheres with charges Q = − q. The spheres are separated by a distance 2a. What is the frequency of small oscillations around the equilibrium point of the bead? You can neglect any friction in the tube.

Hint: When the bead is only slightly displaced, the force acting on it changes negligiblythis is the picture: http://fykos.org/rocnik27/obrazky/s1u5-zadani.png

So, it is better? Do you understand me?

We understand you. Now answer the question in #2.

Numeriprimi said:
Hint: When the bead is only slightly displaced, the force acting on it changes negligibly

So, it is better? Do you understand me?

Yes, like this it works. I thought that the line connecting the two sphere goes along the tube.
Thank you for the clarification.

So, the net force is the vector sum of forces from spheres... Ok? And if the bead isn't between the speheres, vector sum isn't 0, so beat have to oscillate?

Numeriprimi said:
So, the net force is the vector sum of forces from spheres... Ok?

OK.

And if the bead isn't between the speheres, vector sum isn't 0, so beat have to oscillate?

It will feel some force that depends on x. Find the force first, then you can think about the effect of the force.

## 1. What is an oscillation of beat in a tube?

An oscillation of beat in a tube refers to the phenomenon of sound waves reflecting back and forth within a closed tube, resulting in a periodic variation in amplitude known as a beat. This occurs when two sound waves of slightly different frequencies are produced simultaneously.

## 2. How does the length of the tube affect the oscillation of beat?

The length of the tube plays a crucial role in determining the frequency and wavelength of the sound waves produced. A longer tube will have a lower resonant frequency, resulting in a slower oscillation of beat. Conversely, a shorter tube will have a higher resonant frequency and a faster oscillation of beat.

## 3. What is the relationship between the frequency of the beat and the frequency of the two sound waves?

The frequency of the beat is equal to the difference between the frequencies of the two sound waves. For example, if one sound wave has a frequency of 100 Hz and the other has a frequency of 110 Hz, the resulting beat frequency will be 10 Hz.

## 4. How does the amplitude of the beat change over time?

The amplitude of the beat will vary periodically over time, with the maximum amplitude occurring when the two sound waves are in phase (crest to crest or trough to trough) and the minimum amplitude occurring when they are out of phase (crest to trough or trough to crest).

## 5. What are some real-world applications of oscillations of beat in tubes?

Oscillations of beat in tubes have many practical applications, such as in musical instruments like flutes and organ pipes, where the length of the tube can be adjusted to produce different notes. They are also used in physics experiments to demonstrate the principles of sound waves and resonance. Additionally, beat frequencies are used in tuning instruments and in frequency modulation in radio communication.

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