A horizontal platform with an object placed on it is executing $S.H.M$. in the vertical direction. The amplitude of oscillation is $3.92 \times {10^{ - 3}}m$. What must be the least period of these oscillations, so that the object is not detached from the platform
AIIMS 1999, Diffcult
Download our app for free and get started
(a) By drawing free body diagram of object during the downward motion at extreme position, for equilibrium of mass
$mg - R = mA$ ($A = $ Acceleration)
For critical condition $R = 0$
so $mg = mA$
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Two masses, both equal to $100\, g$, are suspended at the ends of identical light strings of length $\lambda = 1.0\, m$, attached to the same point on the ceiling (see figure). At time $t = 0$, they are simultaneously released from rest, one at angle $\theta_1 = 1^o$, the other at angle $\theta_2 = 2^o$ from the vertical. The masses will collide
A particle executes $SHM$ on a straight line path. The amplitude of oscillation is $2\, cm.$ When the displacement of the particle from the mean position is $1\, cm,$ the numerical value of magnitude of acceleration is equal to the numerical value of magnitude of velocity. The frequency of $SHM$ (in $second^{-1}$) 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 $x=$__________.