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Physics

Simple Pendulum

A simple pend...

Question

A simple pendulum of length $2 m$ oscillates such that its speed at the lowest point is $4 m / s$ . It goes to a certain height and reverses its motion. At this position, find the angle that it makes with the vertical.

35^{o}

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45^{o}

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49^{o}

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53^{o}

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Solution

The correct option is D $53^{o}$
Kinetic energy = Potential energy

$\frac{1}{2} m v^{2} = m g h$

$h = \frac{v^{2}}{2 g}$

$L (1 - cos θ) = \frac{v^{2}}{2 g}$

$\Rightarrow cos θ = 1 - \frac{v^{2}}{2 g L} = 1 - \frac{4^{2}}{2 \times 10 \times 2} = \frac{3}{5}$

$\Rightarrow θ = {cos}^{- 1} (\frac{3}{5}) = 53^{o}$

Angle is $53^{o}$

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A simple pendulum of length 2 m oscillates such that its speed at the lowest point is 4 m/s. It goes to a certain height and reverses its motion. At this position, find the angle that it makes with the vertical.

The correct option is D 53oKinetic energy = Potential energy 12mv2=mgh h=v22g L(1−cosθ)=v22g ⇒cosθ=1−v22gL=1−422×10×2=35 ⇒θ=cos−1(35)=53o Angle is 53o

A simple pendulum of length $2 m$ oscillates such that its speed at the lowest point is $4 m / s$ . It goes to a certain height and reverses its motion. At this position, find the angle that it makes with the vertical.