Question

Two light springs of force constant $k_1$ and $k_2$ and a block of mass m are 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 springs is $60$cm. If the block moves along AB with a velocity $120$ cm/sec in between the springs, calculate the period (in sec) of oscillation of the block.
$(k_1=1.8N/m,k_2=3.2N/m,m=200gm)$.

Answer 1

$\text{Solution}$

The mass will strike the right spring, compress it. The K.E. of the mass will convert into P.E. of the spring. Again the spring will return to its natural size thereby verting its P.E. to K.E. of the block. The time taken for this process will be

$\frac{T}{2}$, where $T=2\pi \sqrt{\frac{m}{k}}$.

$\therefore t_1=\frac{T}{2}=\pi \sqrt{\frac{m}{k_2}}=\pi \sqrt{\frac{200}{1000}\left(3.2\right)}=0.785sec$

The block will move from A to B without any acceeleration. The time taken will be

$t_2=\frac{60}{120}=0.5$

Now the block will compress the left spring and then the spring again attains its natural length. The time taken will be oscillation. The time taken for doing so

$t_3=\pi \sqrt{\frac{m}{k_l}}=\pi \sqrt{\frac{0.2}{1.8}}=1.05sec$

Again the block moves from B to A, completion one oscillation. The time taken for doing so

$t_4=\frac{60}{120}=0.5$

$\therefore$ The complete time of oscillation will be $=t_l+t_2+t_3+t_4=0.785+\mathrm{0.51.05}+05=2.83$

$\text{Hence 2.835 is the correct option}$