Question

The block shown in the figure oscillates in simple harmonic motion with amplitude $12\text{cm}$. The amplitude of the point P is :

• A. $6\text{cm}$
• B. $12\text{cm}$
• C. $4\text{cm}$
• D. $2\text{cm}$

Answer 1

The correct option is C $4\text{cm}$

Given that
$A=12\text{cm}$
Spring constant of each spring $=2K$

Equivalent spring constant (for right side of point P)
$\frac{1}{K_{eq}}=\frac{1}{2K}+\frac{1}{2K}$
$\Rightarrow K_{eq}=K$


In last figure, springs are shown in the maximum compressed position.
Let the spring of spring constant $2K$ be compressed by $x_1$ and that of spring constant $K$ be compressed by $x_2$.
So, $x_1+x_2=12\text{cm}$ ...(i)
and at point P, forces in both springs will be same in magnitude.
$2K\left(x_1\right)=K\left(x_2\right)$
$\Rightarrow 2x_1=x_2$
Putting the value of $x_2$ in eq (ii),
$\Rightarrow x_1+2x_1=12$
$\Rightarrow x_1=\frac{12}{3}=4\text{cm}$ will be the amplitude of point P.