$A$ spring-mass system vibrates such that the mass travels on a surface with a coefficient of friction $\mu$. The mass is released after compressing the spring by a distance $a$ and it travels up to a distance $b$ after its equilibrium position. Then,while traveling from $x = -a$ to $x = b$,the reduction in its amplitude will be:

$A$ small block of mass $m$,having charge $q$,is placed on a frictionless inclined plane making an angle $\theta$ with the horizontal. There exists a uniform magnetic field $B$ parallel to the inclined plane but perpendicular to the length of the spring. If $m$ is slightly pulled on the inclined plane in the downward direction and released,the time period of oscillation will be (assume that the block does not leave contact with the plane):

$A$ $5\, kg$ collar is attached to a spring of spring constant $500\, N/m$. It slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by $10\, cm$ and released. The time period of oscillation is

The total mechanical energy of a spring-mass system in simple harmonic motion is $E = \frac{1}{2}m\omega^2 A^2$. Suppose the oscillating particle is replaced by another particle of double the mass while the amplitude $A$ remains the same. The new mechanical energy will:

When a block of mass $m$ is suspended separately by two different springs having time periods $t_1$ and $t_2$,respectively. If the same mass is connected to the series combination of both springs,then its time period is given by:

Two bodies $A$ and $B$ of equal masses are suspended from two separate springs of force constants $k_1$ and $k_2$ respectively. If the two bodies oscillate such that their maximum velocities are equal,the ratio of the amplitudes of oscillation of $A$ and $B$ will be