Question

For a simple pendulum, find the angular amplitude ${\mathit{\theta }}_{\mathbf{m}}$at which the restoring torque required for simple harmonic motion deviates from the actual restoring torque by$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\%}$. (See “Trigonometric Expansions” in Appendix E.)

Short answer

The angular amplitude ${\theta }_m$for a simple pendulum is$14°$.

Step-by-step solution

The given data

• The deviation of restoring torque required for SHM from the actual restoring torque is$1.0\%$.
• Trigonometric functions from Appendix E of the book.

Understanding the concept of SHM

Using the formula for restoring torque required for SHM and actual restoring torque, we can find the angular amplitude for a simple pendulum.

Formula:

Restoring torque required for SHM, $\tau =-L{F}_g\mathrm{sin\theta }$ (i)

Step 3: Calculation for the angular amplitude

Actual restoring torque is,

$\tau =-L{F}_g\theta$ (ii)

So, the deviation of restoring torque required for SHM from the actual restoring torque is given by the difference of equation (i) and (ii) as:

$$\begin{gathered} \frac{\left(\right(-L{F}_g\sin \theta )-(-L{F}_g\theta \left)\right)}{-L{F}_g\sin \theta }=0.01 \\ \frac{\left(\right(\sin \theta )-(\theta \left)\right)}{\sin \theta }=0.01 \\ 1-\frac{\theta }{\sin \theta }=0.01 \end{gathered}$$

We can write for $\mathrm{si}n\theta$from Trigonometric Expansions given in APPENDIX E of the book as:

$\sin \theta =\theta -\frac{{\theta }^3}{6}$

Hence, the angular amplitude of the pendulum is given as:

$$\begin{gathered} 1-\frac{\theta }{\theta -{\displaystyle \frac{{\theta }^3}{6}}}=0.01 \\ 1-\frac{1}{1-{\displaystyle \frac{{\theta }^2}{6}}}=0.01 \\ \frac{1-{\displaystyle \frac{{\theta }^6}{6}}}{{\displaystyle \frac{{\theta }^2}{6}}}=100 \\ \frac{6}{{\theta }^2}=101 \\ \theta =\sqrt{\frac{6}{101}} \\ =0.24\mathrm{rad} \\ =14° \end{gathered}$$

Therefore, the angular amplitude${\theta }_m$ for a simple pendulum is$14°$.