The function $(\sin \omega t - \cos \omega t)$ represents

The motion of a particle executing simple harmonic motion is described by the displacement function,$x(t) = A \cos (\omega t + \phi)$. If the initial $(t = 0)$ position of the particle is $1 \; cm$ and its initial velocity is $\omega \; cm/s$,what are its amplitude and initial phase angle? The angular frequency of the particle is $\pi \; s^{-1}$. If instead of the cosine function,we choose the sine function to describe the $SHM$: $x = B \sin (\omega t + \alpha)$,what are the amplitude and initial phase of the particle with the above initial conditions?

$A$ 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 .........

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 displacement of a particle varies according to the relation $x = 3 \sin 100t + 8 \cos^2 50t$. Which of the following is incorrect about this motion?

The differential equation of a simple harmonic motion is given by $\frac{d^2y}{dt^2} + ky = 0$. What is the time period?