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The variation of kinetic energy $(KE)$ of a particle executing simple harmonic motion with the displacement $(x)$ starting from mean position to extreme position $(A)$ is given by
A mass m is suspended from a spring of length l and force constant $K$. The frequency of vibration of the mass is ${f_1}$. The spring is cut into two equal parts and the same mass is suspended from one of the parts. The new frequency of vibration of mass is ${f_2}$. Which of the following relations between the frequencies is correct
Two particles are executing $SHM$ in a straight line. Amplitude $'A'$ and time period $'T'$ of both the particles are equal. At time $t = 0$ one particle is at displacement $x_1 = +A$ and other at ${x_2} = \frac{{ - A}}{2}$ and they are approaching towards each other. Time after which they will cross each other is
$2$ particles $p$ and $q$ describe $SHM$ of same amplitude $a$ and same frequency $f$ along straight line, the maximum distance between the two particle $a\sqrt 2 $ . The initial phase difference between particle is
Two particles undergo $SHM$ along parallel lines with the same time period $(T)$ and equal amplitudes. At a particular instant, one particle is at its extreme position while the other is at its mean position. They move in the same direction. They will cross each other after a further time
Two simple harmonic motions are represented by equations ${y_1} = 4\,\sin \,\left( {10t + \phi } \right)$ and ${y_2} = 5\,\cos \,10\,t$ What is the phase difference between their velocities?
A vibratory motion is represented by $x = 2\,A\,\cos \,\omega t + A\,\cos \,\left( {\omega t + \frac{\pi }{2}} \right) + A\,\cos \,\left( {\omega t + \pi } \right) + \frac{A}{2}\,\cos \,\left( {\omega t + \frac{{3\pi }}{2}} \right)$ The resultant amplitude of motion is