Two mutually perpendicular simple harmonic vibrations have the same amplitude,frequency,and phase. When they superimpose,the resultant form of vibration will be:

The displacement of a particle in a periodic motion is given by $y = 4 \cos^{2}\left(\frac{t}{2}\right) \sin(1000 t)$. This displacement may be considered as the result of the superposition of $n$ independent harmonic oscillations. Here $n$ is

Two simple harmonic motions $y_1 = A \sin \omega t$ and $y_2 = A \cos \omega t$ are superimposed on a particle of mass $m.$ The total mechanical energy of the particle is:

Two particles are executing simple harmonic motion of the same amplitude $A$ and frequency $\omega$ along the $x$-axis. Their mean positions are separated by a distance $X_0$ $(X_0 > A)$. If the maximum separation between them is $(X_0 + A)$,the phase difference between their motions is:

When two displacements represented by $y_1 = a \sin(\omega t)$ and $y_2 = b \cos(\omega t)$ are superimposed,the motion is

Two simple harmonic motions are represented by the equations $x_{1} = 5 \sin(2 \pi t + \frac{\pi}{4})$ and $x_{2} = 5 \sqrt{2}(\sin 2 \pi t + \cos 2 \pi t)$. The amplitude of the second motion is ....... times the amplitude of the first motion.