Two bodies of masses $1\, kg$ and $4\, kg$ are connected to a vertical spring, as shown in the figure. The smaller mass executes simple harmonic motion of angular frequency $25\, rad/s$, and amplitude $1.6\, cm$ while the bigger mass remains stationary on the ground. The maximum force exerted by the system on the floor is ..... $N$ ( take $g = 10\, ms^{-2}$)

For the damped oscillator shown in Figure the mass mof the block is $200\; g , k=90 \;N m ^{-1}$ and the damping constant $b$ is $40 \;g s ^{-1} .$ Calculate

$(a)$ the period of oscillation,

$(b)$ time taken for its amplitude of vibrations to drop to half of Its inittal value, and

$(c)$ the time taken for its mechanical energy to drop to half its initial value.

A body executes simple harmonic motion under the action of a force $F_1$ with a time period $(4/5)\, sec$. If the force is changed to $F_2$ it executes $SHM$ with time period $(3/5)\, sec$. If both the forces $F_1$ and $F_2$ act simultaneously in the same direction on the body, its time period (in $seconds$ ) is

In the given figure, a mass $M$ is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is $k$. The mass oscillates on a frictionless surface with time period $T$ and amplitude $A$. When the mass is in equilibrium position, as shown in the figure, another mass $m$ is gently fixed upon it. The new amplitude of oscillation will be

A block of mass $m$ attached to massless spring is performing oscillatory motion of amplitude $'A'$ on a frictionless horizontal plane. If half of the mass of the block breaks off when it is passing through its equilibrium point, the amplitude of oscillation for the remaining system become $fA.$ The value of $f$ is

How the period of oscillation depend on the mass of block attached to the end of spring ?