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 }](https://tex.z-dn.net/?f=%5Cfrac%7B2%5Cpi%20%7D%7B%5Comega%20%7D)
Where
is the angular velocity and it is given by,
= ![\sqrt{\frac{k}{m} }](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%20%7D)
Now if the spring constant is doubled then,
![k_{2} = 2k](https://tex.z-dn.net/?f=k_%7B2%7D%20%3D%202k)
Thus,
=![\frac{2\pi }{\sqrt{\frac{2k}{m} } }](https://tex.z-dn.net/?f=%5Cfrac%7B2%5Cpi%20%7D%7B%5Csqrt%7B%5Cfrac%7B2k%7D%7Bm%7D%20%7D%20%7D)
![\frac{T_{2} }{T} = \frac{\frac{2\pi }{\sqrt{\frac{2k}{m} } }}{\frac{2\pi }{\sqrt{\frac{k}{m} } }}](https://tex.z-dn.net/?f=%5Cfrac%7BT_%7B2%7D%20%7D%7BT%7D%20%3D%20%5Cfrac%7B%5Cfrac%7B2%5Cpi%20%7D%7B%5Csqrt%7B%5Cfrac%7B2k%7D%7Bm%7D%20%7D%20%7D%7D%7B%5Cfrac%7B2%5Cpi%20%7D%7B%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%20%7D%20%7D%7D)
![\frac{T_{2} }{T} = \sqrt{\frac{k}{2k} } = \sqrt{\frac{1}{2} }](https://tex.z-dn.net/?f=%5Cfrac%7BT_%7B2%7D%20%7D%7BT%7D%20%3D%20%5Csqrt%7B%5Cfrac%7Bk%7D%7B2k%7D%20%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B1%7D%7B2%7D%20%7D)
![T_{2} = \frac{T}{\sqrt{2} }](https://tex.z-dn.net/?f=T_%7B2%7D%20%3D%20%5Cfrac%7BT%7D%7B%5Csqrt%7B2%7D%20%7D)
Thus, The period would decrease by sqrt(2).
Hence, option D is correct.