A circular arc of mass $m$ is connected with the help of two massless strings as shown in the figuw in vertical plane. About point $P$, small oscillations are given in the plane of the arc. Time period of the oscillations of $SHM$ will be
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Kinetic energy of a particle executing simple harmonic motion in straight line is $pv^2$ and potential energy is $qx^2$, where $v$ is speed at distance $x$ from the mean position. It time period is given by the expression
A body oscillates with a simple harmonic motion having amplitude $0.05\, m .$ At a certain instant, its displacement is $0.01\, m$ and acceleration is $1.0 \,m / s ^{2} .$ The period of oscillation is
A pendulum has time period $T$. If it is taken on to another planet having acceleration due to gravity half and mass $ 9 $ times that of the earth then its time period on the other planet will be
A particle of mass $m$ executes simple harmonic motion with amplitude $a$ and frequency $v$. The average kinetic energy during its motion from the position of equilibrium to the end is
A particle executes a simple harmonic motion of time period $T$. Find the time taken by the particle to go directly from its mean position to half the amplitude
A block of mass $m$ attached to massless spring is performing oscillatory motion of amplitude $'A'$ on a frictionless horizontal plane. If half of the mass of the block breaks off when it is passing through its equilibrium point, the amplitude of oscillation for the remaining system become $fA.$ The value of $f$ is
A simple pendulum oscillates freely between points $A$ and $B$. We now put a peg (nail) at the point $C$ as shown in above figure. As the pendulum moves from $A$ to the right, the string will bend at $C$ and the pendulum will go to its extreme point $D$. Ignoring friction, the point $D$
$2$ particles $p$ and $q$ describe $SHM$ of same amplitude $a$ and same frequency $f$ along straight line, the maximum distance between the two particle $a\sqrt 2 $ . The initial phase difference between particle is