$A$ body executing simple harmonic motion has a maximum acceleration equal to $24 \, m/s^2$ and a maximum velocity equal to $16 \, m/s$. The amplitude of the simple harmonic motion is:

$A$ clock $S$ is based on the oscillation of a spring,and a clock $P$ is based on pendulum motion. Both clocks run at the same rate on Earth. On a planet having the same density as Earth but twice the radius,which of the following is true?

Two particles are in $SHM$ in a straight line. Amplitude $A$ and time period $T$ of both the particles are equal. At time $t=0$,one particle is at displacement $y_1 = +A$ and the other at $y_2 = -A/2$,and they are approaching towards each other. After what time do they cross each other?

$A$ body describes simple harmonic motion with an amplitude of $5\; cm$ and a period of $0.2\; s$. Find the acceleration and velocity of the body when the displacement is $(a)\; 5\; cm$,$(b)\; 3\; cm$,and $(c)\; 0\; cm$.

As shown in the figures,a uniform rod $OO^{\prime}$ of length $l$ is hinged at the point $O$ and held in place vertically between two walls using two massless springs of same spring constant $K$. The springs are connected at the midpoint and at the top-end $(O^{\prime})$ of the rod,as shown in Fig. $1$,and the rod is made to oscillate by a small angular displacement. The frequency of oscillation of the rod is $f_1$. On the other hand,if both the springs are connected at the midpoint of the rod,as shown in Fig. $2$,and the rod is made to oscillate by a small angular displacement,then the frequency of oscillation is $f_2$. Ignoring gravity and assuming motion only in the plane of the diagram,the value of $\frac{f_1}{f_2}$ is:

For a particle executing $S.H.M.$,where $x$ is the displacement from the equilibrium position,$v$ is the velocity at any instant,and $a$ is the acceleration at any instant,then: