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A positively charged object with a mass of \(0.115 \mathrm{kg}\) oscillates at the end of a spring, generating ELF (extremely low frequency) radio waves that have a wavelength of \(4.80 \times 10^{7} \mathrm{m} .\) The frequency of these radio waves is the same as the frequency at which the object oscillates. What is the spring constant of the spring?

Short Answer

Expert verified
The spring constant is approximately 18 N/m.

Step by step solution

01

Calculate Frequency Using Wavelength

First, we calculate the frequency of the ELF radio waves using the speed of light formula. The speed of light is given by \( c = 3 \times 10^8 \text{ m/s} \). Since the radio waves have a wavelength of \( \lambda = 4.80 \times 10^7 \text{ m} \), the frequency \( f \) is calculated as follows:\[ f = \frac{c}{\lambda} = \frac{3 \times 10^8 \text{ m/s}}{4.80 \times 10^7 \text{ m}} = 6.25 \text{ Hz} \].
02

Relate Frequency to Spring Oscillation

The frequency \( f \) of an object oscillating on a spring is related to the mass \( m \) and the spring constant \( k \) by the formula:\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \].We know \( f = 6.25 \text{ Hz} \) and \( m = 0.115 \text{ kg} \).
03

Solve for the Spring Constant

Rearrange the frequency formula to solve for the spring constant \( k \):\[ \sqrt{\frac{k}{m}} = 2\pi f \]Thus,\[ \frac{k}{m} = (2\pi f)^2 \]\[ k = m (2\pi f)^2 \]Substitute the known values:\[ k = 0.115 \text{ kg} \times (2\pi \times 6.25 \text{ Hz})^2 \approx 18 \text{ N/m} \].

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Key Concepts

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

Oscillation Frequency
When an object moves back and forth in a regular rhythm, it is said to be oscillating. This motion is often seen with objects attached to springs. The frequency at which this oscillation occurs is called the **oscillation frequency**. It refers to how many complete cycles the object makes per second and is measured in Hertz (Hz).
  • For this exercise, the frequency of the oscillation is directly related to the frequency of the generated ELF radio waves.
  • The frequency can be calculated if the wavelength of the emitted radio waves is known, using the speed of light formula.
The frequency of an oscillating object suspended on a spring depends on two factors: the mass of the object and the spring constant. This relationship can be expressed through the formula:\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]where:
  • \( f \) is the frequency in Hertz,
  • \( k \) is the spring constant, and
  • \( m \) is the mass of the object.
Thus, by knowing the frequency of the oscillation, we can determine other quantities like the spring constant if the mass is given.
Wavelength of Radio Waves
Radio waves are a type of electromagnetic wave. They travel at the speed of light, which is \( c = 3 \times 10^8 \text{ m/s} \). A key property of any wave, including radio waves, is its **wavelength**.The wavelength is the distance between two identical points on consecutive waves, such as from crest to crest. In this problem, ELF radio waves have a wavelength of \(4.80 \times 10^{7} \text{ m}\). To calculate the frequency of these waves, we must remember the fundamental relationship:\[ f = \frac{c}{\lambda} \]where:
  • \( f \) is the frequency in Hertz,
  • \( c \) is the speed of light, and
  • \( \lambda \) is the wavelength of the wave.
By rearranging this equation, we can find the frequency if the wavelength is known. This formula is instrumental in understanding how radio waves function and interact with their environments.
ELF Radio Waves
**ELF radio waves** stand for "Extremely Low Frequency" radio waves. They have long wavelengths and very low frequencies. In this case, the wavelength of ELF radio waves is so large, \(4.80 \times 10^7 \text{ m}\), making it an excellent match for transmitting over long distances and through water and soil.
  • ELF frequencies are typically in the range of 3–30 Hz, but in this problem, the given frequency is 6.25 Hz.
  • These frequencies are used for applications like communication with submarines and scientific research below the Earth's surface.
Even though their frequency is low, the long wavelength allows ELF waves to penetrate the Earth's surface or the ocean depths, which is why they are valuable in certain specialized communication scenarios.

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Most popular questions from this chapter

A stationary particle of charge \(q=2.6 \times 10^{-8} \mathrm{C}\) is placed in a laser beam (an electromagnetic wave) whose intensity is \(2.5 \times 10^{3} \mathrm{W} / \mathrm{m}^{2}\). Determine the magnitudes of the (a) electric and (b) magnetic forces exerted on the charge. If the charge is moving at a speed of \(3.7 \times 10^{4} \mathrm{m} / \mathrm{s}\) perpendicular to the magnetic field of the electromagnetic wave, find the magnitudes of the (c) electric and (d) magnetic forces exerted on the particle.

An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is \(\bar{S}=1.23 \times 10^{9} \mathrm{W} / \mathrm{m}^{2} .\) What is the rms value of (a) the electric field and (b) the magnetic field in the electromagnetic wave emitted by the laser?

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A truck driver is broadcasting at a frequency of 26.965 MHz with a CB (citizen's band) radio. Determine the wavelength of the electromagnetic wave being used. The speed of light is \(c=2.9979 \times 10^{8} \mathrm{m} / \mathrm{s}\).

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