A ring is suspended from a point $S$ on its rim as shown in the figure. When displaced from equilibrium, it oscillates with time period of $1\,second.$ The radius of the ring is ..... $m$ (take $g = \pi ^2$ )
AIEEE 2012, Diffcult
Download our app for free and get started
We know that the time period of a physical pendulum is given by $T=2 \pi \sqrt{\frac{I_{\text {support }}}{m g l_{\text {cm }}}}$
$I_{\text {support }}=m R^{2}+m R^{2}=2 m R^{2}$
$T=2 \pi \sqrt{\frac{\left(2 m R^{2}\right)}{m g R}} =2 \pi \sqrt{\frac{2 R}{g}}$
$\Rightarrow R=\frac{T^{2} g}{8 \pi^{2}}$
A seconds pendulum is a pendulum whose period is precisely $two\, seconds; one \,second$ for a swing in one direction and $one\, second$ for the return swing. Ie. $T=2 \varepsilon,$ so we have
$R=\frac{2^{2} \times \pi^{2}}{8 \pi^{2}}=0.5 \mathrm{m}$
Download our appand get started for free
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.*
Similar Questions
- 1The variation of the acceleration $a$ of the particle executing $S.H.M.$ with displacement $x$ is as shown in the figureView Solution
- 2The amplitude of the vibrating particle due to superposition of two $SHMs,$View Solution
$y_1 = \sin \left( {\omega t + \frac{\pi }{3}} \right)$ and $y_2 = \sin \omega t$ is :
- 3A particle executes simple harmonic motion. Its amplitude is $8 \,cm$ and time period is $6 \,s$. The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude, is ............. $s$View Solution
- 4Two simple harmonic motions of angular frequency $100$ and $1000\,\,rad\,s^{-1}$ have the same displacement amplitude. The ratio of their maximum acceleration isView Solution
- 5A spring mass system preforms $S.H.M.$ If the mass is doubled keeping amplitude same, then the total energy of $S.H.M.$ will become :View Solution
- 6Two 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 collideView Solution
- 7A potential is given by $V(x)=k(x+a)^2 / 2$ for $x < 0$ and $V(x)=k(x-a)^2 / 2$ for $x > 0$. The schematic variation of oscillation period $T$ for a particle performing periodic motion in this potential as a function of its energy $E$ isView Solution
- 8A small block is connected to one end of a massless spring of un-stretched length $4.9 \ m$. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by $0.2$ $m$ and released from rest at $t =0$. It then executes simple harmonic motion with angular frequency $\omega=\frac{\pi}{3} \ rad / s$.View Solution
Simultaneously at $t=0$, a small pebble is projected with speed $v$ from point $P$ at an angle of $45^{\circ}$ as shown in the figure. Point $P$ is at a horizontal distance of $10 \ cm$ from $O$. If the pebble hits the block at $t=1 \ s$, the value of $v$ is (take $g =10 \ m / s ^2$ )
- 9The function ${\sin ^2}(\omega t)$ representsView Solution
- 10The vertical extension in a light spring by a weight of $1\, kg$ suspended from the wire is $9.8\, cm$. The period of oscillationView Solution