Two particles, 1 and 2, each of mass $m$, are connected by a massless spring, and are on a horizontal frictionless plane, as shown in the figure. Initially, the two particles, with their center of mass at $x_0$, are oscillating with amplitude $a$ and angular frequency $\omega$. Thus, their positions at time $t$ are given by $x_1(t)=\left(x_0+d\right)+a \sin \omega t$ and $x_2(t)=\left(x_0-d\right)-a \sin \omega t$, respectively, where $d>2 a$. Particle 3 of mass $m$ moves towards this system with speed $u_0=a \omega / 2$, and undergoes instantaneous elastic collision with particle 2 , at time $t_0$. Finally, particles 1 and 2 acquire a center of mass speed $v_{\mathrm{cm}}$ and oscillate with amplitude $b$ and the same angular frequency $\omega$.

On a frictionless horizontal plane, a bob of mass $m=0.1 \mathrm{~kg}$ is attached to a spring with natural length $l_{0}=0.1 \mathrm{~m}$. The spring constant is $k_{1}=0.009 \,\mathrm{Nm}^{-1}$ when the length of the spring $l>l_{0}$ and is $k_{2}=0.016 \,\mathrm{Nm}^{-1}$ when $l < l_{0}$. Initially the bob is released from $l=$ $0.15 \mathrm{~m}$. Assume that Hooke's law remains valid throughout the motion. If the time period of the full oscillation is $T=(n \pi) s$, then the integer closest to $n$ is __________.

^{-7}m

^{2}. If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad s

^{−1}. If the Young’s modulus of the material of the wire is n $$ \times $$ 10

^{9}Nm

^{-2}, the value of n is