(B) By the principle of conservation of mechanical energy,the potential energy at the initial position must equal the potential energy at the final po

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$\cos^{-1}\left[ \frac{L\cos \alpha - l}{L - l} \right]$

(B) By the principle of conservation of mechanical energy,the potential energy at the initial position must equal the potential energy at the final position.Let the pivot point be the reference level for potential energy $(PE = 0)$.Initial height of the mass $M$ from the pivot is $h_i = -L \cos \alpha$.Final height of the mass $M$ from the pivot after hitting the pin is $h_f = -l - (L - l) \cos \theta$.Equating the potential energies: $Mg(-L \cos \alpha) = Mg(-l - (L - l) \cos \theta)$.$-L \cos \alpha = -l - (L - l) \cos \theta$.$(L - l) \cos \theta = L \cos \alpha - l$.$\cos \theta = \frac{L \cos \alpha - l}{L - l}$.Therefore,$\theta = \cos^{-1}\left[ \frac{L \cos \alpha - l}{L - l} \right]$.

Class 11 Physics Oscillations SHM of Simple Pendulum

Difficult

$A$ simple pendulum consisting of a mass $M$ attached to a string of length $L$ is released from rest at an angle $\alpha$. $A$ pin is located at a distance $l$ below the pivot point. When the pendulum swings down,the string hits the pin as shown in the figure. The maximum angle $\theta$ which the string makes with the vertical after hitting the pin is:

A
$\cos^{-1}\left[ \frac{L\cos \alpha + l}{L - l} \right]$
B
$\cos^{-1}\left[ \frac{L\cos \alpha - l}{L - l} \right]$
C
$\cos^{-1}\left[ \frac{L\cos \alpha + l}{L + l} \right]$
D
$\cos^{-1}\left[ \frac{L\cos \alpha - l}{L + l} \right]$

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SHM of Simple Pendulum

$A$ small sphere oscillates simple harmonically in a watch glass whose radius of curvature is $1.6 \ m$. The period of oscillation of the sphere is (acceleration due to gravity $g = 10 \ m/s^2$) (in $\pi \ s$)

EasyMHT CET 2024

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The bob of a simple pendulum having length $l$ is displaced from the mean position to an angular position $\theta$ with respect to the vertical. If it is released,then the velocity of the bob at the equilibrium position is:

Difficult

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$A$ simple pendulum of time period $1\,s$ and length $l$ is hung from a fixed support at $O$. The bob is at a distance $H$ vertically above $A$ on the ground. The amplitude is $\theta_0$. The string snaps at $\theta = \frac{\theta_0}{2}$. Find the time taken by the bob to hit the ground and the distance from $A$ where the bob hits the ground. Assume $\theta_0$ to be small,so that $\sin \theta_0 \approx \theta_0$ and $\cos \theta_0 \approx 1$.

Difficult

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$A$ pendulum is suspended by a string of length $250\,cm$. The mass of the bob of the pendulum is $200\,g$. The bob is pulled aside until the string is at $60^{\circ}$ with the vertical,as shown in the figure. After releasing the bob,the maximum velocity attained by the bob will be . . . . . . $m/s$. (if $g = 10\,m/s^2$)

MediumJEE MAIN 2022

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$A$ block of mass $m$ is sliding on a fixed frictionless concave surface of radius $R$. It is released from rest at point $P$ which is at a height of $H \ll R$ from the lowest point $Q$.
$(a)$ What is the potential energy as a function of $\theta$,taking the lowest point $Q$ as the reference level for potential energy?
$(b)$ What is the kinetic energy as a function of $\theta$?
$(c)$ What is the time taken for the particle to reach from point $P$ to the lowest point $Q$?
$(d)$ How much force is exerted by the block on the concave surface at the point $Q$?

DifficultKVPY 2010

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$A$ small sphere oscillates simple harmonically in a watch glass whose radius of curvature is $1.6 \ m$. The period of oscillation of the sphere is (acceleration due to gravity $g = 10 \ m/s^2$) (in $\pi \ s$)

The bob of a simple pendulum having length $l$ is displaced from the mean position to an angular position $\theta$ with respect to the vertical. If it is released,then the velocity of the bob at the equilibrium position is:

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