$A$ particle executing $SHM$ has a maximum speed of $0.5 \ m s^{-1}$ and a maximum acceleration of $1.0 \ m s^{-2}$. The angular frequency of oscillation is

  • A
    $2 \ rad \ s^{-1}$
  • B
    $0.5 \ rad \ s^{-1}$
  • C
    $2 \pi \ rad \ s^{-1}$
  • D
    $0.5 \pi \ rad \ s^{-1}$

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Similar Questions

The amplitude and the time period in a $S.H.M.$ are $0.5\, cm$ and $0.4\, s$ respectively. If the initial phase is $\pi/2$ radian,then the equation of $S.H.M.$ will be:

Out of the following functions representing the motion of a particle,which represent $SHM$?
$(A)\; y = \sin \omega t - \cos \omega t$
$(B)\; y = \sin^3 \omega t$
$(C)\; y = 5 \cos \left( \frac{3\pi}{4} - 3\omega t \right)$
$(D)\; y = 1 + \omega t + \omega^2 t^2$

The displacement $x$ (in metre) of a particle in simple harmonic motion is related to time $t$ (in seconds) as $x = 0.01 \cos \left( \pi t + \frac{\pi}{4} \right)$. The frequency of the motion will be

The motion of a particle varies with time according to the relation $y = a(\sin \omega t + \cos \omega t)$.

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When a particle of mass $m$ moves on the $x$-axis in a potential of the form $V(x)=kx^2$,it performs simple harmonic motion. The corresponding time period is proportional to $\sqrt{\frac{m}{k}}$,as can be seen easily using dimensional analysis. However,the motion of a particle can be periodic even when its potential energy increases on both sides of $x=0$ in a way different from $kx^2$ and its total energy is such that the particle does not escape to infinity. Consider a particle of mass $m$ moving on the $x$-axis. Its potential energy is $V(x)=\alpha x^4$ $(\alpha>0)$ for $|x|$ near the origin and becomes a constant equal to $V_0$ for $|x| \geq X_0$ (see figure).
$1.$ If the total energy of the particle is $E$,it will perform periodic motion only if
$(A)$ $E < 0$
$(B)$ $E > 0$
$(C)$ $V_0 > E > 0$
$(D)$ $E > V_0$
$2.$ For periodic motion of small amplitude $A$,the time period $T$ of this particle is proportional to
$(A)$ $A \sqrt{\frac{m}{\alpha}}$
$(B)$ $\frac{1}{A} \sqrt{\frac{m}{\alpha}}$
$(C)$ $A \sqrt{\frac{\alpha}{m}}$
$(D)$ $A \sqrt{\frac{\alpha}{m}}$
$3.$ The acceleration of this particle for $|x|>X_0$ is
$(A)$ proportional to $V_0$
$(B)$ proportional to $\frac{V_0}{mX_0}$
$(C)$ proportional to $\sqrt{\frac{V_0}{mX_0}}$
$(D)$ zero
Give the answer for questions $1, 2$ and $3$.

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