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
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A particle free to move along the $x-$axis has potential energy given by $U(x) = k[1 - \exp {( - x)^2}]$ for $ - \infty \le x \le + \infty $, where k is a positive constant of appropriate dimensions. Then
The displacement of an oscillating particle varies with time (in seconds) according to the equation $y (cm) = sin \frac{\pi }{2}\left( {\frac{t}{2} + \frac{1}{3}} \right)$. The maximum acceleration of the particle is approximately ..... $cm/s^2$
The amplitude of a damped oscillator decreases to $0.9\,times$ its original magnitude in $5\,s.$ In another $10\,s$ it will decrease to $\alpha $ times its original magnitude, where $\alpha $ equals
A particle of mass $m$ in a unidirectional potential field have potential energy $U(x)=\alpha+2 \beta x^2$, where $\alpha$ and $\beta$ are positive constants. Find its time period of oscillation.
Two pendulums have time periods $T$ and $\frac{{5T}}{4}.$They start $S.H.M.$ at the same time from the mean position. What will be the phase difference between them after the bigger pendulum has complete one oscillation ..... $^o$
Two, spring $P$ and $Q$ of force constants $k_p$ and ${k_Q}\left( {{k_Q} = \frac{{{k_p}}}{2}} \right)$ are stretched by applying forces of equal magnitude. If the energy stored in $Q$ is $E$, then the energy stored in $P$ is
A simple harmonic oscillator has an amplitude $A$ and time period $6 \pi$ second. Assuming the oscillation starts from its mean position, the time required by it to travel from $x=A$ to $x=\frac{\sqrt{3}}{2} A$ will be $\frac{\pi}{x}$ s, where $x=$__________.
$Assertion :$ The time-period of pendulum, on a satellite orbiting the earth is infinity.
$Reason :$ Time-period of a pendulum is inversely proportional to $\sqrt g$