B M Sharma Solutions for Chapter: Linear And Angular Simple Harmonic Motion, Exercise 5: DPP

The displacement of a particle varies with time as, $x=12 \sin \left(\omega t\right)-16{\sin }^3\left(\omega t\right)$ (in $\mathrm{cm}$). If its motion is SHM, then, its maximum acceleration is,

A particle is acted simultaneously by mutually perpendicular simple harmonic motion $x=a \cos \omega t$ and $y=a \sin \omega t$. The trajectory of motion of the particle will be,

A disc of radius $R$ and mass $M$ is pivoted at the rim and is set for small oscillations. If a simple pendulum has to have the same period as that of the disc, the length of the simple pendulum should be,

A particle of mass $m=2 \mathrm{kg}$ executes SHM in $xy$-plane between points $A$ and $B$ under the action of force, $\overrightarrow{F}=F_x\hat{i}+F_y\hat{j}$. The minimum time taken by the particle to move from $A$ to $B$ is $1 \mathrm{s}$. At $t=0$, the particle is at $x=2$ and $y=2$. Then, $F_x$ as function of time $t$ is,

The displacement of a particle varies with time according to the relation, $y=a \sin \omega t+b \cos \omega t$.

Three simple harmonic motions in the same direction having the same amplitude and same period are superimposed. If each differ in phase from the next by $45°$, then,

A function $x=A{\sin }^2\omega t+B{\cos }^2\omega t+C\sin \omega t \cos \omega t$ represents SHM,

The position of a particle wrt to the origin varies according to the relation, $x=3\sin 100t+8 {\cos }^{2}50t$. Which of the following is/are correct about this motion?