$A$ mass $m$ oscillates with simple harmonic motion with frequency $f = \frac{\omega}{2\pi}$ and amplitude $A$ on a spring with constant $K$. Therefore:

$A$ 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:

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:

$A$ clock $S$ is based on the oscillations 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,then:

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:

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