A block of mass 4 kg is executing simple harmonic motion on a...

The initial frequency of the block is given by:
$\nu = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{m}}$

When the putty of mass $5\text{ kg}$ is dropped on the block of mass $4\text{ kg}$, the new total mass becomes $M = 4 + 5 = 9\text{ kg}$.
The new frequency $\nu'$ is:
$\nu' = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{M}} = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{9}}$

Taking the ratio of the new frequency to the old frequency:
$\dfrac{\nu'}{\nu} = \sqrt{\dfrac{m}{M}} = \sqrt{\dfrac{4}{9}} = \dfrac{2}{3}$
$\Rightarrow \nu' = \dfrac{2}{3}\nu$

At the mean position, the velocity of the block is maximum. Let the initial velocity be $v$.
$v = A\omega = A(2\pi\nu)$

Since the putty is dropped vertically, there is no external force in the horizontal direction. By conservation of linear momentum along the horizontal direction:
$mv = Mv'$
$4v = 9v' \Rightarrow v' = \dfrac{4}{9}v$

The new maximum velocity $v'$ is related to the new amplitude $A'$ and new angular frequency $\omega'$ by:
$v' = A'\omega'$

Substituting $v' = \dfrac{4}{9}v$ and $\omega' = 2\pi\nu' = \dfrac{2}{3}\omega$:
$A'\left(\dfrac{2}{3}\omega\right) = \dfrac{4}{9}(A\omega)$

Solving for $A'$:
$A' = \dfrac{4}{9} \times \dfrac{3}{2} A = \dfrac{2}{3}A$

Thus, the new frequency is $\dfrac{2}{3}\nu$ and the new amplitude is $\dfrac{2}{3}A$.

Answer: $\nu' = \dfrac{2}{3}\nu$, $A' = \dfrac{2}{3}A$