$A$ body of mass $5\, g$ is executing $S.H.M.$ about a point with amplitude $10\, cm$. Its maximum velocity is $100\, cm/s$. Its velocity will be $50\, cm/s$ at a distance of:

$A$ body is executing $S.H.M.$ When its displacement from the mean position is $4 \, cm$ and $5 \, cm$,the corresponding velocity of the body is $10 \, cm/sec$ and $8 \, cm/sec$. Then the time period of the body is

$A$ particle is executing simple harmonic motion with amplitude $A$. At a distance $x$ from the mean position,when the particle is moving towards the extreme position,it receives a blow in the direction of motion which instantaneously doubles its velocity. What is the new amplitude of the particle? (Frequency remains constant during the motion)

$A$ particle is performing simple harmonic motion about $x = 0$ with an amplitude $a$ and periodic time $T$. The speed of the particle at $x = \frac{a}{3}$ will be

$A$ particle performing $SHM$ is found at its equilibrium position at $t = 1 \, s$. It is found to have a speed of $0.25 \, m/s$ at $t = 2 \, s$. If the period of oscillation is $6 \, s$,calculate the amplitude of oscillation.

$A$ particle executing simple harmonic motion starts from the mean position with amplitude $A$ and periodic time $T$. At what displacement is its speed one-fourth of the maximum speed?