A simple pendulum with bob of mass m and length l is held in position at an angle - with the vertical- Its speed when it passes the lowest position is

We will use law of conservation of energy nbsp; K.E. of the pendulum at point B is zero P.E. of the pendulum at point 'B' nbsp; P.E. = mgh Let us find h C ...

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Standard IX

Physics

Law of Conservation of Energy

A simple pend...

Question

A simple pendulum with bob of mass m and length $l$ is held in position at an angle $θ$ with the vertical. Its speed when it passes the lowest position is

\sqrt{g l c o s θ}

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\sqrt{2 g l c o s θ}

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g l (1 - c o s θ)

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\sqrt{2 g l (1 - c o s θ)}

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Solution

The correct option is D $\sqrt{2 g l (1 - c o s θ)}$

We will use law of conservation of energy
K.E. of the pendulum at point B is zero
P.E. of the pendulum at point 'B'
P.E. = mgh
Let us find h
Consider the right angled triangle BCD

In $Δ B C D \Rightarrow \frac{B a s e}{H e i g h t} = c o s θ From triangle we get C D = l c o s θ$
From fig we can see that h = AD
Hence we will find AD
$A D = A C - C D = l - l c o s θ A D = l (1 - c o s θ) h = A D = l (1 - c o s θ) P.E at point 'B'= m g l (1 - c o s θ) At point A, K.E=P.E at point B Law of conservation of energy \frac{1}{2} m v^{2} = m g l (1 - c o s θ) v^{2} = 2 g l (1 - c o s θ) v = \sqrt{2 g l (1 - c o s θ)}$

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A simple pendulum with bob of mass m and length l is held in position at an angle θ with the vertical. Its speed when it passes the lowest position is

A simple pendulum with bob of mass m and length $l$ is held in position at an angle $θ$ with the vertical. Its speed when it passes the lowest position is