Consider a block of mass `'m'` attached to a spring of spring constant `'K'` as shown. Now a block of same mass `'m'` moving with velocity `'u'` strikes the other block and stick to it. Both blocks collectively start performing `SHM`.

The amplitude `(A)` of their `SHM` is

Options

A. `(mg)/(K)`
B. `(2mg)/(K)`
C. `sqrt((mmu^(2))/(2K)+(m^(2)g^(2))/(K^(2)))`
D. `sqrt((mmu^(2))/(2K)-(m^(2)g^(2))/(K^(2)))`

Answer

C

Explanation

`v = omegasqrt(A^(2)-x^(2))`
`u/2 = sqrt((K)/(2m)) sqrt(A^(2) - ((mg)/(K))^(2)`
`(u^(2))/(4) = K/(2m) xx (A^(2) - (m^(2)g^(2))/(K^(2)))`
`A = sqrt((mu^(2))/(2K)+(m^(2)g^(2))/(K^(2)))`