Two springs with negligible masses and force constants $K_1 = 200\, Nm^{-1}$ and $K_2 = 160\, Nm^{-1}$ are attached to a block of mass $m = 10\, kg$ as shown in the figure. Initially,the block is at rest at the equilibrium position where both springs are neither stretched nor compressed. At time $t = 0$,a sharp impulse of $50\, Ns$ is given to the block with a hammer.

  • A
    Period of oscillations for the mass $m$ is $\frac{\pi}{3} \, s.$
  • B
    Maximum velocity of the mass $m$ during its oscillation is $5\, ms^{-1}.$
  • C
    Data are insufficient to determine maximum velocity.
  • D
    $(A)$ and $(B)$ both.

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$A$ small block is connected to one end of a massless spring of un-stretched length $4.9 \ m$. The other end of the spring is fixed at $O$. The system lies on a horizontal frictionless surface. The block is stretched by $0.2 \ m$ and released from rest at $t = 0$. It then executes simple harmonic motion with angular frequency $\omega = \frac{\pi}{3} \ rad/s$. Simultaneously at $t = 0$,a small pebble is projected with speed $v$ from point $P$ at an angle of $45^{\circ}$ as shown in the figure. Point $P$ is at a horizontal distance of $10 \ m$ from $O$. If the pebble hits the block at $t = 1 \ s$,the value of $v$ is (take $g = 10 \ m/s^2$):

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 $x_1(t) = (x_0 + d) + a \sin \omega t$ and $x_2(t) = (x_0 - d) - a \sin \omega t$,respectively,where $d > 2a$. Particle $3$ of mass $m$ moves towards this system with speed $u_0 = a \omega / 2$ and undergoes an instantaneous elastic collision with particle $2$ at time $t_0$. Finally,particles $1$ and $2$ acquire a center of mass speed $v_{cm}$ and oscillate with amplitude $b$ and the same angular frequency.
$(1)$ If the collision occurs at time $t_0 = 0$,the value of $v_{cm} / (a \omega)$ will be
$(2)$ If the collision occurs at time $t_0 = \pi / (2 \omega)$,then the value of $4b^2 / a^2$ will be

Phase space diagrams are useful tools in analyzing all kinds of dynamical problems. They are especially useful in studying the changes in motion as initial position and momentum are changed. Here we consider some simple dynamical systems in one-dimension. For such systems, phase space is a plane in which position is plotted along the horizontal axis and momentum is plotted along the vertical axis. The phase space diagram is the $x(t)$ vs. $p(t)$ curve in this plane. The arrow on the curve indicates the time flow. For example, the phase space diagram for a particle moving with constant velocity is a straight line as shown in the figure. We use the sign convention in which position or momentum upwards (or to the right) is positive and downwards (or to the left) is negative.
$1.$ The phase space diagram for a ball thrown vertically up from the ground is:
$2.$ The phase space diagram for simple harmonic motion is a circle centered at the origin. In the figure, the two circles represent the same oscillator but for different initial conditions, and $E_1$ and $E_2$ are the total mechanical energies respectively. Then:
$(A) E_1 = \sqrt{2} E_2$
$(B) E_1 = 2 E_2$
$(C) E_1 = 4 E_2$
$(D) E_1 = 16 E_2$
$3.$ Consider the spring-mass system, with the mass submerged in water, as shown in the figure. The phase space diagram for one cycle of this system is:
Give the answer for questions $1, 2,$ and $3.$

The center of a disk of radius $r$ and mass $m$ is attached to a spring of spring constant $k$,inside a ring of radius $R > r$ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring,without slipping. The spring can only be stretched or compressed along the periphery of the ring,following Hooke's law. In equilibrium,the disk is at the bottom of the ring. Assuming small displacement of the disc,the time period of oscillation of the center of mass of the disk is written as $T = \frac{2 \pi}{\omega}$. The correct expression for $\omega$ is ($g$ is the acceleration due to gravity):

The displacement-time graph of a particle executing $SHM$ is shown. Which of the following statements is wrong?

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