Question

If the spring constant is doubled, what value does the period have for a mass on a spring? A. The period would double by square `sqrt(2)`. B. The period would be halved by `sqrt(2)`. C. The period would increase by `sqrt(2)`. D. The period would decrease by `sqrt(2)`.

Answer 1

Answer:

Answer D. is INCORRECT on PLATO!!!!

Explanation:

Answer 2

Answer:

The period would decrease by sqrt(2)

Explanation:

The restoring force is given by,

F = -kx

According to Newton's second law of motion,

ma = -kx

ma + kx = 0

The time period is given by,

T =$\frac{2\pi }{\omega }$

Where $\omega$ is the angular velocity and it is given by,

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

Now if the spring constant is doubled then,

$k_{2} = 2k$

Thus,

$T_{2}$ =$\frac{2\pi }{\sqrt{\frac{2k}{m} } }$

$\frac{T_{2} }{T} = \frac{\frac{2\pi }{\sqrt{\frac{2k}{m} } }}{\frac{2\pi }{\sqrt{\frac{k}{m} } }}$

$\frac{T_{2} }{T} = \sqrt{\frac{k}{2k} } = \sqrt{\frac{1}{2} }$

$T_{2} = \frac{T}{\sqrt{2} }$

Thus, The period would decrease by sqrt(2).

Hence, option D is correct.