What is the phase difference between the velocity and displacement of a particle executing $SHM$ (in $^{\circ}$)?

  • A
    $0$
  • B
    $90$
  • C
    $180$
  • D
    $270$

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$A$ particle executes $S.H.M.$ according to the equation $x=10 \cos \left[2 \pi t+\frac{\pi}{2}\right] \, cm$,where $t$ is in seconds. The magnitude of the velocity of the particle at $t=\frac{1}{6} \, s$ will be .............. $cm/s$.

The equation of a particle executing simple harmonic motion is given by $x = \sin \pi (t + 1/3) \, m$. At $t = 1 \, s$,the speed of the particle will be .......... $cm \, s^{-1}$ (Given: $\pi = 3.14$)

The displacement of a particle executing $SHM$ is given by $x = 3 \sin \left(2 \pi t + \frac{\pi}{4}\right)$,where $x$ is in $m$ and $t$ is in $s$. The amplitude and maximum speed of the particle are:

$A$ 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.

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If a simple pendulum oscillates with an amplitude of $50 \,mm$ and time period of $2 \,s$, then its maximum velocity is (in $\,ms^{-1}$)

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