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11-32P

Expert-verifiedFound in: Page 292

Book edition
7th

Author(s)
Douglas C. Giancoli

Pages
978 pages

ISBN
978-0321625922

**Derive a formula for the maximum speed \({v_{\max }}\) of a simple pendulum bob in terms of g, the length l, and the maximum angle of swing \({\theta _{\max }}\).**

The formula for the maximum speed is \({v_{\max }} = \sqrt {2gl\left( {1 - \cos {\theta _{\max }}} \right)} \).

**The simple pendulum is said to be in simple harmonic motion when it oscillates back and forth, and the effect of friction can be ignored. **

**The speed of the pendulum bob is maximum when it moves through its equilibrium position.**

** **

The maximum angle of swing \({\theta _{\max }}\).

The maximum speed is \({v_{{\rm{max}}}}\).

The length of the pendulum is \(l\).

A simple pendulum of length *l* consisting of a bob of mass *m *at the maximum angle of swing \({\theta _{\max }}\) is shown in the figure below:

Take the lowest point of the bob as the reference level (*h* = 0).

The total energy of the pendulum bob at the lowest point (\(\theta = 0\)) where its speed is maximum \(\left( {{v_{\max }}} \right)\)is given as:

\(\begin{aligned}{c}{E_{{\rm{bottom}}}} &= K{E_{{\rm{bottom}}}} + P{E_{{\rm{bottom}}}}\\ &= \frac{1}{2}mv_{\max }^2 + mg(0)\\ &= \frac{1}{2}mv_{\max }^2 + 0\end{aligned}\)

Here, \(K{E_{{\rm{bottom}}}}\) and \(P{E_{{\rm{bottom}}}}\) are the kinetic and potential energy at the bottom, m is the mass and g is the gravitational acceleration.

The total energy of the pendulum bob at the highest point (at top where \(\theta \) is maximum) is given as:

\(\begin{aligned}{c}{E_{{\rm{top}}}} &= K{E_{{\rm{top}}}} + P{E_{{\rm{top}}}}\\ &= 0 + mgh\end{aligned}\)

Here, \(K{E_{{\rm{top}}}}\) and \(P{E_{{\rm{top}}}}\) are the kinetic and potential energy at the top.

According to the law of conservation of energy, the total energy at the top \(\left( {{E_{{\rm{top}}}}} \right)\)must be equal to the total energy at the bottom \(\left( {{E_{{\rm{bottom}}}}} \right)\). Thus,

\(\begin{aligned}{c}{E_{{\rm{top}}}} &= {E_{{\rm{bottom}}}}\\(0 + mgh) &= \left( {\frac{1}{2}mv_{\max }^2 + 0} \right)\\mgh &= \frac{1}{2}mv_{\max }^2\\{v_{\max }} = \sqrt {2gh} \end{aligned}\) ….. (i)

Thus, the height of the bob at the top point is:

\(h = \left( {l - l\cos {\theta _{\max }}} \right)\)

Substitute this value of *h* in equation (i).

\(\begin{aligned}{c}{v_{\max }} &= \sqrt {2gh} \\ &= \sqrt {2gl\left( {1 - \cos {\theta _{\max }}} \right)} \end{aligned}\)

This is the required formula for the maximum velocity of the pendulum bob.

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