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A particle executing a simple harmonic motion of period $2\ s$ . When it is at its extreme displacement from its mean position, it receives an additional energy equal to what it had in its mean position. Due to this , in its subsequent motion,
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 particle is executing $S.H.M.$ with total mechanical energy $90 \,J$ and amplitude $6 \,cm$. If its energy is somehow decreased to $40 \,J$ then its amplitude will become ........ $cm$
Two particles P and Q describe S.H.M. of same amplitude $a$, same frequency $f$ along the same straight line. The maximum distance between the two particles is a $\sqrt{2}$.The initial phase difference between the particle is -
A particle executes $S.H.M.$ and its position varies with time as $x=A$ sin $\omega t$. Its average speed during its motion from mean position to mid-point of mean and extreme position is
Two bodies performing $SHM$ have same amplitude and frequency. Their phases at a certain instant are as shown in the figure. The phase difference between them is
A particle of mass $10$ grams is executing simple harmonic motion with an amplitude of $0.5\, m$ and periodic time of $(\pi /5)$ seconds. The maximum value of the force acting on the particle is ... $N$