A particle executes simple harmonic oscillation with an amplitude $a.$ The period of oscillation is $T.$ The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is
A$\frac{T}{8}$
B$\;\frac{T}{{12}}$
C$\;\frac{T}{2}$
D$\;\frac{T}{4}$
AIPMT 2007, Medium
Download our app for free and get started
B$\;\frac{T}{{12}}$
b $x(t)=a \sin \omega t$ $(from \,the\, equilibrium\, position)$
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.*
A vertical mass spring system executes simple harmonic oscillations with a period of $2\,s$. A quantity of this system which exhibits simple harmonic variation with a period of $1\, sec$ is
Infinite springs with force constant $k$, $2k$, $4k$ and $8k$.... respectively are connected in series. The effective force constant of the spring will be
A solid cylinder of density $\rho_0$, cross-section area $A$ and length $l$ floats in a liquid of density $\rho\left( >\rho_0\right)$ with its axis vertical, as shown. If it is slightly displaced downward and released, the time period will be .......
A clock $S$ is based on oscillations of a spring and a clock $P$ is based on pendulum motion. Both clocks run at the same rate on earth. On a planet having same density as earth but twice the radius then
The displacement of a particle moving in $S.H.M.$ at any instant is given by $y = a\sin \omega t$. The acceleration after time $t = \frac{T}{4}$ is (where $T$ is the time period)
A pendulume clock loses $12\;s$ a day if the temperature is $40^oC$ and gains $4\;s$ a day if the temperature is $20^oC$. The temperature at which the clock will show correct time, and the coeffecient of linear expansion $(\alpha)$ of the metal of the pendulum shaft are respectively