Question 14.11: Simple pendulum with damping. A simple pendulum has a length......
Simple pendulum with damping. A simple pendulum has a length of 1.0 \mathrm{~m} (Fig. 14-22). It is set swinging with small-amplitude oscillations. After 5.0 minutes, the amplitude is only 50 \% of what it was initially. (a) What is the value of \gamma for the motion? (b) By what factor does the frequency, f^{\prime}, differ from f, the undamped frequency?
APPROACH We assume the damping force is proportional to angular speed, d \theta / d t. The equation of motion for damped harmonic motion is
x=A e^{-\gamma t} \cos \omega^{\prime} t, \quad \text { where } \quad \gamma=\frac{b}{2 m} \quad \text { and } \quad \omega^{\prime}=\sqrt{\frac{k}{m}-\frac{b^{2}}{4 m^{2}}}
for motion of a mass on the end of a spring. For the simple pendulum without damping, we saw in Section 14-5 that
F=-m g \theta
for small \theta. Since F=m a, where a can be written in terms of the angular acceleration \alpha=d^{2} \theta / d t^{2} as a=\ell \alpha=\ell d^{2} \theta / d t^{2}, then F=m \ell d^{2} \theta / d t^{2}, and
\ell \frac{d^{2} \theta}{d t^{2}}+g \theta=0
Introducing a damping term, b(d \theta / d t), we have
\ell \frac{d^{2} \theta}{d t^{2}}+b \frac{d \theta}{d t}+g \theta=0
which is the same as Eq. 14-15 with \theta replacing x, and \ell and g replacing m and k.
m \frac{d^{2} x}{d t^{2}}+b \frac{d x}{d t}+k x=0 (14-15)
(a) We compare Eq. 14-15 with our equation just above and see that our equation x=A e^{-\gamma t} \cos \omega^{\prime} t becomes an equation for \theta with
\gamma=\frac{b}{2 \ell} \text { and } \quad \omega^{\prime}=\sqrt{\frac{g}{\ell}-\frac{b^{2}}{4 \ell^{2}}}
At t=0, we rewrite Eq. 14-16 with \theta replacing x as
x=A e^{-\gamma t} \cos \omega^{\prime} t (14-16)
\theta_{0}=A e^{-\gamma \cdot 0} \cos \omega^{\prime} \cdot 0=A
Then at t=5.0 \mathrm{~min}=300 \mathrm{~s}, the amplitude given by Eq. 14-16 has fallen to 0.50 \mathrm{~A}, so
0.50 A=A e^{-\gamma(300 \mathrm{~s})}
We solve this for \gamma and obtain \gamma=\ln 2.0 /(300 \mathrm{~s})=2.3 \times 10^{-3} \mathrm{~s}^{-1}.
(b) We have \ell=1.0 \mathrm{~m}, so b=2 \gamma \ell=2\left(2.3 \times 10^{-3} \mathrm{~s}^{-1}\right)(1.0 \mathrm{~m})=4.6 \times 10^{-3} \mathrm{~m} / \mathrm{s}. Thus \left(b^{2} / 4 \ell^{2}\right) is very much less than g / \ell\left(=9.8 \mathrm{~s}^{-2}\right), and the angular frequency of the motion remains almost the same as that of the undamped motion. Specifically (see Eq. 14-20),
f^{\prime}=\frac{\omega^{\prime}}{2 \pi}=\frac{1}{2 \pi} \sqrt{\frac{k}{m}-\frac{b^{2}}{4 m^{2}}} (14-20)
f^{\prime}=\frac{1}{2 \pi} \sqrt{\frac{g}{\ell}}\left[1-\frac{\ell}{g}\left(\frac{b^{2}}{4 \ell^{2}}\right)\right]^{\frac{1}{2}} \approx \frac{1}{2 \pi} \sqrt{\frac{g}{\ell}}\left[1-\frac{1}{2} \frac{\ell}{g}\left(\frac{b^{2}}{4 \ell^{2}}\right)\right]
where we used the binomial expansion. Then, with f=(1 / 2 \pi) \sqrt{g / \ell} (Eq. 14-12b),
f=\frac{\omega}{2 \pi}=\frac{1}{2 \pi} \sqrt{\frac{g}{\ell}}, \qquad [ \theta \rm small] (14-12b)
\frac{f-f^{\prime}}{f} \approx \frac{1}{2} \frac{\ell}{g}\left(\frac{b^{2}}{4 \ell^{2}}\right)=2.7 \times 10^{-7}
So f^{\prime} differs from f by less than one part in a million.