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Crazy boy [7]
3 years ago
6

A positively charged objectwith a mass of 0.114 kg oscillates at the end of a spring, generating ELF (extremely low frequency) r

adio waves that have a wavelength of 4.92 x 107 m. The frequency of these radio waves is the same as the frequency at which the object oscillates. What is the sprinE constant of the
Physics
1 answer:
katen-ka-za [31]3 years ago
3 0

Answer:

  • k = 167.33 N/m

Explanation:

  • The radio waves have a fixed relationship between the propagation speed (the speed of light in vacuum), the frequency and the wavelength, as follows:
  • v = c = λ*f

        where c= speed of light in vacuum = 3*10⁸ m/s, λ = wavelength =  

        4.92*10⁷ m.

        Solving for f, we get the frequency of the radio waves:

        f = 6.1 Hz

  • Now, from the Hooke's law, we know that the mass attached at the end of the spring oscillates with an angular frequency defined by  a fixed relationship between the spring constant k and the mass m, as follows:

       \omega_{o}^{2} =\frac{k}{m}  (1)

  • Now, we know that there exists a fixed relationship between the angular frequency and the frequency, as follows:

       \omega = 2*\pi *f (2)    

  • We also know that f in (2) is the same that we got for the radio waves, so replacing (2) in (1), and rearranging terms, we can solve for k, as follows:
  • k = 4*\pi ^{2}*f^{2} *m = 4*\pi ^{2} * (6.1Hz)^{2} * 0.114 kg = 167.33 N/m

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