$A$ simple harmonic motion is represented by $F(t) = 10\sin(20t + 0.5)$. The amplitude of the $S.H.M.$ is $a = $ ....

The function $\sin^2(\omega t)$ represents

The displacement of a particle executing simple harmonic motion is given by
$y = A_{0} + A \sin \omega t + B \cos \omega t$
Then the amplitude of its oscillation is given by

If the displacement $(x)$ and velocity $(v)$ of a particle executing simple harmonic motion are related through the expression $4v^2 = 25 - x^2$, then the time period is

$A$ particle executes simple harmonic motion according to the equation $4 \frac{d^2 x}{d t^2} + 320 x = 0$. Its time period of oscillation is .........

$A$ particle performs $SHM$ on the $x-$axis with a time period of $0.5 \, s,$ such that its velocity is zero at $x = -3 \, cm$ and at $x = 9 \, cm$. It was located at $x = 0$ and moving in the negative $x-$direction at $t = 0$. The equation of $SHM$ of the particle is: