A simple pendulum with a bob of mass m oscillates from A to C and back to A such that P B is H. If the acceleration due to gravity is g, then the velocity of the bob as it passes through B is

Question

A simple pendulum with a bob of mass m oscillates from A to C and back to A such that P B is H. If the acceleration due to gravity is g, then the velocity of the bob as it passes through B is (A) mgH (B) √2gH (C) zero (D) 2gH

Tardigrade Admin · Accepted Answer

Potential energy at A (or C) = Kinetic energy at B. Thus (1/2) mvB2 = mg H or vb = √ 2 g H

Q. A simple pendulum with a bob of mass $m$ oscillates from $A$ to $C$ and back to $A$ such that $PB$ is $H$ . If the acceleration due to gravity is $g$ , then the velocity of the bob as it passes through $B$ is

$m g H$

$2 g H$

zero

$2 g H$

Potential energy at $A$ (or $C$ ) = Kinetic energy at $B$ .
Thus $\frac{1}{2} m v_{B}^{2} = m g H$
or $v_{b} = 2 g H$

Q. A simple pendulum with a bob of mass m oscillates from A to C and back to A such that PB is H. If the acceleration due to gravity is g, then the velocity of the bob as it passes through B is

mgH

2gH​

zero

2gH

Potential energy at A (or C) = Kinetic energy at B. Thus 21​mvB2​=mgH or vb​=2gH​

Q. A simple pendulum with a bob of mass $m$ oscillates from $A$ to $C$ and back to $A$ such that $PB$ is $H$ . If the acceleration due to gravity is $g$ , then the velocity of the bob as it passes through $B$ is

$m g H$

$2 g H$

$2 g H$

Potential energy at $A$ (or $C$ ) = Kinetic energy at $B$ .
Thus $\frac{1}{2} m v_{B}^{2} = m g H$
or $v_{b} = 2 g H$