Question

If the spring constant is doubled, what value does the period have for a mass on a spring?

Answer 1

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

$\omega = \sqrt{\frac{k}{m}}$

(this is obtained by solving the differential equation $m\ddot{{ x}}-kx=0$)

If k doubles, i.e., k'=2k, then

$\omega'=\sqrt{\frac{k'}{m}}=\sqrt{\frac{2k}{m}}=\sqrt{2}\sqrt{\frac{k}{m}}=\omega\sqrt{2}$

Since the angular frequency is $\omega = \frac{2\pi}{T}$, we can say that

$\omega\sqrt{2}=\frac{2\pi}{T}\sqrt{2}=\frac{2\pi}{\frac{T}{\sqrt{2}}}=\frac{2\pi}{T'}$

and so it becomes clear that the period T will decrease by sqrt(2) as stated in choice (D).

Answer 2

Answer:

D. The period would decrease by `sqrt(2)`. is wrong on clever/plato

Explanation:

i got it wrong :(