Answer:
The period would decrease by `sqrt(2)`.
Explanation:
The angular frequency omega of an oscillating mass m due to a spring with a constant k is given by
[tex]\omega = \sqrt{\frac{k}{m}}[/tex]
(this is obtained by solving the differential equation [tex]m\ddot{{ x}}-kx=0[/tex])
If k doubles, i.e., k'=2k, then
[tex]\omega'=\sqrt{\frac{k'}{m}}=\sqrt{\frac{2k}{m}}=\sqrt{2}\sqrt{\frac{k}{m}}=\omega\sqrt{2}[/tex]
Since the angular frequency is [tex]\omega = \frac{2\pi}{T}[/tex], we can say that
[tex]\omega\sqrt{2}=\frac{2\pi}{T}\sqrt{2}=\frac{2\pi}{\frac{T}{\sqrt{2}}}=\frac{2\pi}{T'}[/tex]
and so it becomes clear that the period T will decrease by sqrt(2) as stated in choice (D).
