Answer:
(c) 4M
Explanation:
The system is a loaded spring. The period of a loaded spring is given by
[tex]T = 2\pi\sqrt{\dfrac{m}{k}}[/tex]
m is the mass and k is the spring constant.
It follows that, since k is constant,
[tex]T\propto\sqrt{m}[/tex]
[tex]\dfrac{T}{\sqrt{m}} = C[/tex] where C represents a constant.
[tex]\dfrac{T_1}{\sqrt{m_1}} = \dfrac{T_2}{\sqrt{m_2}}[/tex]
[tex]m_2 = m_1\left(\dfrac{T_2}{T_1}\right)^2[/tex]
When the period is doubled, [tex]T_2 = 2T_1[/tex].
[tex]m_2 = m_1\left(\dfrac{2T_1}{T_1}\right)^2 = 4m_1[/tex]
Hence, the mass is replaced by 4M.
For doubling the period, the object should be replaced with one of mass that should be option c. 4M.
Since the system should be considered as the loaded spring.
So
T = 2π√m/k
Here
m be the mass
k means the spring constant
Now
T1 ÷ √m1 = T2 ÷ √m2
In the case when the period is doubled
So,
T2 = 2T1
So,
m2 = m1 = (2T1/T1)^2 = 4m1
= 4m
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