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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
The velocity of a particle executing SHM varies with displacement $( x )$ as $4 v ^2=50- x ^2$. The time period of oscillations is $\frac{x}{7} s$. The value of $x$ is $............$ $\left(\right.$ Take $\left.\pi=\frac{22}{7}\right)$
A body of mass $1\,kg$ is executing simple harmonic motion. Its displacement $y(cm)$ at $t$ seconds is given by $y = 6\sin (100t + \pi /4)$. Its maximum kinetic energy is ..... $J$
A body of mass $0.01 kg$ executes simple harmonic motion $(S.H.M.)$ about $x = 0$ under the influence of a force shown below : The period of the $S.H.M.$ is .... $s$
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 ${45^o}$, then
The amplitude of a simple pendulum, oscillating in air with a small spherical bob, decreases from $10\, cm$ to $8\, cm$ in $40\, seconds$ . Assuming that Stokes law is valid, and ratio of the coefficient of viscosity of air to that of carbon dioxide is $1.3$ . The time in which amplitude of this pendulum will reduce from $10\, cm$ to $5\, cm$ in carbon dioxide will be close to ..... $s$ $(ln\, 5 = 1.601,ln\, 2 = 0 .693)$