The equation of $SHM$ is given as:
$x = 3 \sin(20\pi t) + 4 \cos(20\pi t)$,
where $x$ is in $cm$ and $t$ is in $seconds$. The amplitude is ..... $cm$.

Two particles $P$ and $Q$ perform $SHM$ with the same amplitude $a$ and frequency $v$ along the same straight line. The maximum distance between the two particles is $a \sqrt{2}$. The initial phase difference between the particles is:

Three simple harmonic motions in the same direction having the same amplitude $a$ and same period are superposed. If each differs in phase from the next by $45^\circ$,then:

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 $S.H.M.$ with the same amplitude of $20 \, cm$ and the same period along the same line about the same equilibrium position. The maximum distance between the two is $20 \, cm$. Their phase difference in radians is equal to

On the superposition of two harmonic oscillations represented by $x_1 = a \sin(\omega t + \phi_1)$ and $x_2 = a \sin(\omega t + \phi_2)$,a resulting oscillation with the same time period and amplitude is obtained. The value of $\phi_1 - \phi_2$ is .... $^o$