A simple pendulum with bob of mass m and length x is held in position at an angle θ1 and then angle θ2 with the vertical. When released from these positions, speeds with which it passes the lowest positions are v1 v2 respectively. Then, (v1/v2) is

Question

A simple pendulum with bob of mass m and length x is held in position at an angle θ1 and then angle θ2 with the vertical. When released from these positions, speeds with which it passes the lowest positions are v1 v2 respectively. Then, (v1/v2) is (A) (1- cos θ1/1- cos θ2) (B) √(1- cos θ1/1- cos θ2) (C) √(2 g x(1- cos θ1)/1- cos θ2) (D) √(1- cos θ1/2 g x(1- cos θ2))

Tardigrade Admin · Accepted Answer

Ui+ki=Uf+kf (Mechanical energy conservation) m g l(1- cos θ)=(1/2) m v2 (v12/v22)=(1- cos θ1/1- cos θ2) (v1/v2)=√(1- cos θ1/1- cos θ2)

Q. A simple pendulum with bob of mass $m$ and length $x$ is held in position at an angle $θ_{1}$ and then angle $θ_{2}$ with the vertical. When released from these positions, speeds with which it passes the lowest positions are $v_{1}$ & $v_{2}$ respectively. Then, $\frac{v _{1}}{v _{2}}$ is

$\frac{1 - c o s θ _{1}}{1 - c o s θ _{2}}$

$\frac{1 - c o s θ _{1}}{1 - c o s θ _{2}}$

$\frac{2 gx ( 1 - c o s θ _{1} )}{1 - c o s θ _{2}}$

$\frac{1 - c o s θ _{1}}{2 gx ( 1 - c o s θ _{2} )}$

$U_{i} + k_{i} = U_{f} + k_{f}$ (Mechanical energy conservation)
$m g l (1 - cos θ) = \frac{1}{2} m v^{2}$
$\frac{v _{1}^{2}}{v _{2}^{2}} = \frac{1 - c o s θ _{1}}{1 - c o s θ _{2}}$
$\frac{v _{1}}{v _{2}} = \frac{1 - c o s θ _{1}}{1 - c o s θ _{2}}$

Q. A simple pendulum with bob of mass m and length x is held in position at an angle θ1​ and then angle θ2​ with the vertical. When released from these positions, speeds with which it passes the lowest positions are v1​ & v2​ respectively. Then, v2​v1​​ is

1−cosθ2​1−cosθ1​​

1−cosθ2​1−cosθ1​​​

1−cosθ2​2gx(1−cosθ1​)​​

2gx(1−cosθ2​)1−cosθ1​​​

Ui​+ki​=Uf​+kf​ (Mechanical energy conservation) mgl(1−cosθ)=21​mv2 v22​v12​​=1−cosθ2​1−cosθ1​​ v2​v1​​=1−cosθ2​1−cosθ1​​​

Q. A simple pendulum with bob of mass $m$ and length $x$ is held in position at an angle $θ_{1}$ and then angle $θ_{2}$ with the vertical. When released from these positions, speeds with which it passes the lowest positions are $v_{1}$ & $v_{2}$ respectively. Then, $\frac{v _{1}}{v _{2}}$ is

$\frac{1 - c o s θ _{1}}{1 - c o s θ _{2}}$

$\frac{1 - c o s θ _{1}}{1 - c o s θ _{2}}$

$\frac{2 gx ( 1 - c o s θ _{1} )}{1 - c o s θ _{2}}$

$\frac{1 - c o s θ _{1}}{2 gx ( 1 - c o s θ _{2} )}$

$U_{i} + k_{i} = U_{f} + k_{f}$ (Mechanical energy conservation)
$m g l (1 - cos θ) = \frac{1}{2} m v^{2}$
$\frac{v _{1}^{2}}{v _{2}^{2}} = \frac{1 - c o s θ _{1}}{1 - c o s θ _{2}}$
$\frac{v _{1}}{v _{2}} = \frac{1 - c o s θ _{1}}{1 - c o s θ _{2}}$