$A$ particle performs simple harmonic motion with amplitude $A$. Its speed is increased to three times at an instant when its displacement is $\frac{2A}{3}$. The new amplitude of motion is $\frac{nA}{3}$. The value of $n$ is . . . . . . .

$A$ particle in $SHM$ is described by the displacement equation $x(t) = A\cos(\omega t + \theta)$. If the initial $(t = 0)$ position of the particle is $1 \, cm$ and its initial velocity is $\pi \, cm/s$,what is its amplitude? The angular frequency of the particle is $\pi \, s^{-1}$.

The instantaneous displacement of a particle in $S.H.M.$ is $x = A \cos \left(\omega t + \frac{\pi}{4}\right)$. The time at which the velocity is maximum for the first time is

$A$ particle executes simple harmonic motion with an amplitude of $4 \, cm$. At the mean position,the velocity of the particle is $10 \, cm/s$. The distance of the particle from the mean position when its speed becomes $5 \, cm/s$ is

$A$ mass attached to a spring performs $S.H.M.$ whose displacement is $x = 3 \times 10^{-3} \cos(2 \pi t) \text{ m}$. The time taken to obtain maximum speed for the first time is

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: $\pi = 3.14$)