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

Step-by-step explanation:

Given

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

wavelength
`\left ( \lambda \right )=3.86* 10^7`

We know frequency =
`(c)/(\lambda )=(3* 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=(1)/(2\pi )\sqrt{(k)/(m)}`

`7.772=(1)/(2\pi )\sqrt{(k)/(0.129)}`

`7.772* 2* \pi =\sqrt{(k)/(0.129)}`

`k=307.70\approx 308 N/m`