Question

Two light springs of force constant $K_1$ and $K_2$ and a block of mass $m$ are arranged in one line $AB$ on a smooth horizontal table such that one end of each spring is fixed on rigid supports and the other end is free as shown in the figure. The distance $CD$ between the free ends of the spring is $60\text{cm}$. If the block moves along $AB$ with a velocity $120\text{cm/s}$ in between the springs, calculate the period of oscillation of the block.
(Assume that the dimensions of the cube is much less than $60\text{cm}$)
$(K_1=1.8\text{N/m},K_2=3.2\text{N/m},m=200\text{gm})$

• A. $1.41\text{s}$
• B. $2.83\text{s}$
• C. $5.64\text{s}$
• D. $1.92\text{s}$

Answer 1

The correct option is B $2.83\text{s}$

Between $C$ and $D$ block will move with constant speed of $120\text{cm/s}$.
Period of oscillation will be (starting from $C$).
$T=t_{CD}+\frac{T_2}{2}+t_{DC}+\frac{T_1}{2}$
$\Rightarrow$The block will travel the distance $CD$ back and forth, and when it comes in contact with spring free end then it will become part of spring-block system on each side for the half time period.
Here, $T_1=2\pi \sqrt{\frac{m}{K_1}}$ and $T_2=2\pi \sqrt{\frac{m}{K_2}}$
$t_{CD}=t_{DC}=\frac{60}{120}=0.5\text{s}$
Given: $(K_1=1.8\text{N/m},K_2=3.2\text{N/m},m=200\text{gm}=0.2\text{kg})$
$\Rightarrow T=0.5+\frac{2\pi }{2}\sqrt{\frac{0.2}{3.2}}+0.5+\frac{2\pi }{2}\sqrt{\frac{0.2}{1.8}}$
$\therefore T\simeq 2.83\text{s}$