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Expert-verified Found in: Page 138 ### Physics Principles with Applications

Book edition 7th
Author(s) Douglas C. Giancoli
Pages 978 pages
ISBN 978-0321625922 # The length of a simple pendulum is 0.72 m, the pendulum bob has a mass of 295 g, and it is released at an angle of $$12^\circ$$ to the vertical. Assume SHM. (a) With what frequency does it oscillate? (b) What is the pendulum bob’s speed when it passes through the lowest point of the swing? (c) What is the total energy stored in this oscillation assuming no losses?

(a) The frequency oscillates at $$f = 0.59\;{\rm{Hz}}$$.

(b) The pendulum bob’s speed when it passes through the lowest point of the swing is $$0.56\;{\rm{m/s}}$$.

(c) The total energy stored in this oscillation assuming no losses is $$4.5 \times {10^{ - 2}}\;{\rm{J}}$$.

See the step by step solution

## Concept of frequency of SHM

In this problem, for calculating the frequency of the oscillation, apply the relation of the frequency with the gravitational acceleration and length of the pendulum.

## Given data

The length of a simple pendulum is $$l = 0.72\;{\rm{m}}$$.

The mass of the bob is $$m = 295\;{\rm{g}}$$.

The initial angle of the bob with the length is $$\theta = 12^\circ$$.

## Calculation of frequency

The frequency of a simple pendulum is:

$$f = \frac{1}{{2\pi }}\sqrt {\frac{g}{l}}$$

Substitute the known values in the above relation to find frequency.

\begin{aligned}f &= \frac{1}{{2\pi }}\sqrt {\frac{{9.8\;{\rm{m/}}{{\rm{s}}^2}}}{{0.72\;{\rm{m}}}}} \\ &= 0.587\;{\rm{Hz}}\\\approx {\rm{0}}{\rm{.59}}\;{\rm{Hz}}\end{aligned}

Hence, the frequency of the simple pendulum is $$0.59\;{\rm{Hz}}$$.

## Calculation of velocity of bob

An illustrative figure is shown below: The relation according to the conservation of energy is given by,

$$K{E_1} + P{E_1} = K{E_2} + P{E_2}$$

Here, kinetic energy of the bob initially is zero, $$K{E_1} = 0$$, as well as the potential energy at the lowest point $$P{E_2} = 0$$, since it will carry momentum due to moving downwards due to gravity.

On plugging the values in the above relation.

\begin{aligned}{c}0 + mg\left( {l - l\cos \theta } \right) &= \frac{1}{2}mv_2^2 + 0\\{v_2} &= \sqrt {2g\left( {l - l\cos \theta } \right)} \\{v_2} &= \sqrt {2\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {0.72\;{\rm{m}} - \left( {0.72\;{\rm{m}}} \right)\cos 12^\circ } \right)} \\{v_2} &= 0.56\;{\rm{m/s}}\end{aligned}

Hence, the velocity of the bob at its lowest point is $$0.56\;{\rm{m/s}}$$.

## Calculation of total energy

Total energy at its lowest point is equal to the kinetic energy at that point, since there will not be any potential energy left at the lowest point. Thus,

\begin{aligned}{c}{E_{\rm{T}}} = K{E_2} + P{E_2}\\ = \frac{1}{2}mv_2^2 + 0\\ = \frac{1}{2}mv_2^2\end{aligned}

Substitute the known values in the above relation to find total energy.

\begin{aligned}{c}{E_{\rm{T}}} &= \frac{1}{2}\left( {295\;{\rm{g}} \times \frac{{1\;{\rm{kg}}}}{{1000\;{\rm{g}}}}} \right){\left( {0.56\;{\rm{m/s}}} \right)^2}\\ &= 4.5 \times {10^{ - 2}}\;{\rm{J}}\end{aligned}

Hence, the total energy at its lowest point is $$4.5 \times {10^{ - 2}}\;{\rm{J}}$$. ### Want to see more solutions like these? 