A spring is stretched by $5 \,\mathrm{~cm}$ by a force $10 \,\mathrm{~N}$. The time period of the oscillations when a mass of $2 \,\mathrm{~kg}$ is suspended by it is :(in $s$)
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Two pendulums of length $121\,cm$ and $100\,cm$ start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is :
A body is executing $S.H.M.$ When its displacement from the mean position is $4\, cm$ and $5\, cm$, the corresponding velocity of the body is $10 \,cm/sec$ and $8\, cm/sec$. Then the time period of the body is
The angular frequency of the damped oscillator is given by, $\omega = \sqrt {\left( {\frac{k}{m} - \frac{{{r^2}}}{{4{m^2}}}} \right)} $ where $k$ is the spring constant, $m$ is the mass of the oscillator and $r$ is the damping constant. If the ratio $\frac{{{r^2}}}{{mk}}$ is $8\%$, the change in time period compared to the undamped oscillator is approximately as follows
$Assertion :$ The time-period of pendulum, on a satellite orbiting the earth is infinity.
$Reason :$ Time-period of a pendulum is inversely proportional to $\sqrt g$
Two particles oscillating in $SHM$ along two very close parallel path such that they have same mean position. The equation of $SHM$ of two particles are $x_1 = A\, sin\,\omega t$ and $x_2 = A\,sin(\omega t + \phi )$ respectively. If maximum distance between them is $\frac{6A}{5}$ then $\phi $ equal to ..... $^o$