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A body of mass $5\, gm$ is executing $S.H.M.$ about a point with amplitude $10 \,cm$. Its maximum velocity is $100\, cm/sec$. Its velocity will be $50\, cm/sec$ at a distance
A loaded vertical spring executes $S.H.M.$ with a time period of $4\; sec$. The difference between the kinetic energy and potential energy of this system varies with a period of ........$sec$
A $LCR$ circuit behaves like a damped harmonic oscillator. Comparing it with a physical springmass damped oscillator having damping constant $\mathrm{b}$, the correct equivalence would be:
The variation of displacement with time of a particle executing free simple harmonic motion is shown in the figure. The potential energy ${U}({x})$ versus time $({t})$ plot of the particle is correctly shown in figure:
A particle is excuting a simple harmonic motion. Its maximum acceleration is $\alpha $ and maximum velocity is $\beta $. Then its frequency of vibration will be
The amplitude of vibration of a particle is given by ${a_m} = ({a_0})/(a{\omega ^2} - b\omega + c);$ where ${a_0},a,b$ and $c$ are positive. The condition for a single resonant frequency is
pendulum made of a uniform wire of cross sectional area $A$ has time period $T$. When an additional mass $M$ is added to its bob, the time period changes to $T_M$. If the Young's modulus of the material of the wire is $Y$ then $\frac{1}{Y}$ is equal to : ($g$ = gravitational acceleration)
The time period of oscillations of a simple pendulum is $1$ minute. If its length is increased by $44 \%$. then its new time period of oscillation will be ......... $s$