Question

A positively-charged object with a mass of 0.129 kg oscillates at the end of a spring, generating ELF (extremely low frequency) radio waves that have a wavelength of 3.86 × 10^7 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?

Answer 1

Answer:308 N/m

Explanation:

Given

mass$\left ( m\right )=0.129 kg$

wavelength$\left ( \lambda \right )=3.86\times 10^7$

We know frequency =$\frac{c}{\lambda }=\frac{3\times 10^8}{3.86\tmes 10^7}$

f=7.772 Hz

As the frequency of radio waves is same as the frequency at which object oscillates

$f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$

$7.772=\frac{1}{2\pi }\sqrt{\frac{k}{0.129}}$

$7.772\times 2\times \pi =\sqrt{\frac{k}{0.129}}$

$k=307.70\approx 308 N/m$