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 =[tex]\frac{2\pi }{\omega }[/tex]
Where [tex]\omega[/tex] is the angular velocity and it is given by,
[tex]\omega[/tex] = [tex]\sqrt{\frac{k}{m} }[/tex]
Now if the spring constant is doubled then,
[tex]k_{2} = 2k[/tex]
Thus,
[tex]T_{2}[/tex] =[tex]\frac{2\pi }{\sqrt{\frac{2k}{m} } }[/tex]
[tex]\frac{T_{2} }{T} = \frac{\frac{2\pi }{\sqrt{\frac{2k}{m} } }}{\frac{2\pi }{\sqrt{\frac{k}{m} } }}[/tex]
[tex]\frac{T_{2} }{T} = \sqrt{\frac{k}{2k} } = \sqrt{\frac{1}{2} }[/tex]
[tex]T_{2} = \frac{T}{\sqrt{2} }[/tex]
Thus, The period would decrease by sqrt(2).
Hence, option D is correct.
