The motion of a simple pendulum excuting $S.H.M$. is represented by following equation.
$Y = A \sin (\pi t +\phi)$, where time is measured in $second$.
The length of pendulum is .............$cm$
JEE MAIN 2022, Medium
Download our app for free and get started
$\omega=\sqrt{\frac{ g }{\ell}}=\pi$
$\frac{ g }{\ell}=\pi^{2} \Rightarrow \ell=\frac{ g }{\pi^{2}}$
$\ell=\frac{980}{\pi^{2}} \approx 99.4 \,cm$
Download our app
and 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.*
Two, spring $P$ and $Q$ of force constants $k_p$ and ${k_Q}\left( {{k_Q} = \frac{{{k_p}}}{2}} \right)$ are stretched by applying forces of equal magnitude. If the energy stored in $Q$ is $E$, then the energy stored in $P$ is
A block is resting on a piston which executes simple harmonic motion with a period $2.0 \,s$. The maximum velocity of the piston, at an amplitude just sufficient for the block to separate from the piston is .......... $ms ^{-1}$
Three masses $700g, 500g$, and $400g$ are suspended at the end of a spring a shown and are in equilibrium. When the $700g$ mass is removed, the system oscillates with a period of $3$ seconds, when the $500 \,gm$ mass is also removed, it will oscillate with a period of ...... $s$
In an angular $SHM$ angular amplitude of oscillation is $\pi $ $rad$ and time period is $0.4\,sec$ then calculate its angular velocity at angular displacement $ \pi/2 \,rad$. ..... $rad/sec$
A spring whose unstretched length is $\ell $ has a force constant $k$. The spring is cut into two pieces of unstretched lengths $\ell_1$ and $\ell_2$ where, $\ell_1 = n\ell_2$ and $n$ is an integer. The ratio $k_1/k_2$ of the corresponding force constants, $k_1$ and $k_2$ will be
A spring executes $SHM$ with mass of $10\,kg$ attached to it. The force constant of spring is $10\,N/m$.If at any instant its velocity is $40\,cm/sec$, the displacement will be .... $m$ (where amplitude is $0.5\,m$)
Two simple pendulum whose lengths are $1\,m$ and $121\, cm$ are suspended side by side. Their bobs are pulled together and then released. After how many minimum oscillations of the longer pendulum will the two be in phase again
lfa simple pendulum has significant amplitude (up to a factor of $1/e$ of original) only in the period between $t = 0\ s$ to $t = \tau \ s$, then $\tau$ may be called the average life of the pendulum. When the spherical bob of the pendulum suffers a retardation ( due to viscous drag) proportional to its velocity with $b$ as the constant of proportionality, the average life time of the pendulum is (assuming damping is small) in seconds
If $x, v$ and $a$ denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period $T$, then, which of the following does not change with time?