In a science museum, a 110 kg brass pendulum bob swings at the end of a 13.2 m -long wire. The pendulum is started at exactly 8:00 a.m. every morning by pulling it 1.7 m to the side and releasing it. Because of its compact shape and smooth surface, the pendulum's damping constant is only 0.010kg/s.

**A. At exactly 12:00 noon, how many oscillations will the pendulum have completed?**

**B. And what is its amplitude?**

### Answer

**(A)** The expression of time period of the pendulum is,

`T=2 \pi \sqrt{\frac{l}{g}}`

The expression of the frequency is,

`f=\frac{1}{T}`

Substitute `2 \pi \sqrt{\frac{l}{g}}`

for `T`

.

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

Substitute `13.2 \mathrm{~m}`

for `\mathrm{l}`

and `9.8 \mathrm{~m} / \mathrm{s}^{2}`

for `\mathrm{g}`

to find frequency.

`f=\frac{1}{2 \pi} \sqrt{\frac{9.8 \mathrm{~m} / \mathrm{s}^{2}}{13.2 \mathrm{~m}}}`

`=0.137 oscillation/ \mathrm{s}`

The time interval between 8:00 AM to 12.00 PM is,

`t=4 \mathrm{~h}`

The number of oscillations completed in time t is, `N=f t`

Substitute `0.137 oscillation/s`

for `f`

and `4 \mathrm{~h}`

for `t`

.

`N=(0.137 oscillation/ \mathrm{s})\left(4 \mathrm{~h}\left(\frac{3600 \mathrm{~s}}{1 \mathrm{~h}}\right)\right)`

`=1973 oscillation`

**The number of oscillations of the pendulum is 1973**

**(B)** The expression of the amplitude at time `t`

is,

`x(t)=A \exp \left(-\frac{b t}{2 m}\right)`

Substitute `110 \mathrm{~kg}`

for `\mathrm{m}, 1.7 \mathrm{~m}`

for `\mathrm{A}, 0.010 \mathrm{~kg} / \mathrm{s}`

for `\mathrm{b}`

, and `4 \mathrm{~h}`

for `\mathrm{t}`

.

`x(t) =(1.7 \mathrm{~m}) \exp \left(-\frac{(0.010 \mathrm{~kg} / \mathrm{s})\left(4 \mathrm{~h}\left(\frac{3600 \mathrm{~s}}{1 \mathrm{~h}}\right)\right)}{2(110 \mathrm{~kg})}\right)`

`=0.883 \mathrm{~m}\left(\frac{10^{-2} \mathrm{~m}}{1 \mathrm{~m}}\right)`

`=88.3 \mathrm{~cm}`

**The amplitude is 88.3 \mathrm{~cm}.**