Answer:
The block's mass should be [tex]3m[/tex]
Explanation:
Given:
Cart with mass [tex]m[/tex]
From the conservation of energy before mass is added,
[tex]\frac{1}{2} mv^{2} = \frac{1}{2} kA^{2}[/tex]
Where [tex]A =[/tex] amplitude of spring mass system, [tex]k =[/tex] spring constant
[tex]A = v\sqrt{\frac{m}{k} }[/tex]
Now new mass [tex]M[/tex] is added to the system,
[tex]\frac{1}{2} (m +M ) v^{2} = \frac{1}{2} k A^{2}[/tex]
[tex]A = v \sqrt{\frac{m +M }{k} }[/tex]
Here, given in question frequency is reduced to half so we can write,
[tex]f' = \frac{f}{2}[/tex]
Where [tex]f =[/tex] frequency of system before mass is added, [tex]f' =[/tex] frequency of system after mass is added.
[tex]\omega ' = \frac{\omega}{2}[/tex]
[tex]\sqrt{\frac{k}{m +M} } = \frac{\sqrt{\frac{k}{m} } }{2}[/tex]
[tex]\frac{k}{m +M } = \frac{k}{4m}[/tex]
[tex]M = 3m[/tex]
Therefore, the block's mass should be [tex]3m[/tex]
