$A$ particle executes $S.H.M.$ with a period of $6 \, s$ and amplitude of $3 \, cm$. Its maximum speed in $cm/s$ is

Maximum speed of a particle in simple harmonic motion is $v_{max}$. Then,the average speed of a particle in one complete oscillation is equal to:

The displacement of a particle executing $SHM$ is given by $x = 10 \sin \left(\omega t + \frac{\pi}{3}\right) \ m$. The time period of motion is $3.14 \ s$. The velocity of the particle at $t = 0$ is . . . . . . $m/s$.

The time period of a particle in simple harmonic motion is $8 \ s$. At $t=0$,it is at the mean position. The ratio of the distances travelled by it in the first and second seconds is

If the maximum velocity and maximum acceleration of a particle executing simple harmonic motion are respectively $5 \,m \,s^{-1}$ and $10 \,m \,s^{-2}$, then the time period of oscillation of the particle is

Two identical pendulums oscillate with a constant phase difference $\frac{\pi}{4}$ and same amplitude. If the maximum velocity of one is $v$,the maximum velocity of the other will be ........