(Solved): Can someone explain since Wiley can't seem to put the effort in? A metronome consists of a small we ...

Can someone explain since Wiley can't seem to put the effort in?

A metronome consists of a small weight that slides on a thin rod; markings show where to put the weight for various tempos. The metronome clicks every time the rod passes the central position. For a tempo of 108 beats per minute, how far above the pivot point should the weight be placed? \( \mathrm{cm} \) (Note: Mechanical metronomes use a double-weighted pendulum, which allows them to be much shorter.) Solution The period of a pendulum is calculated \( T=2 \pi \sqrt{\frac{L}{g}} \). Given that the metronome is moving at 108 beats per minute, which is 2 beats per period, we have \( T=\frac{2 \cdot 60 \mathrm{~s}}{108}=1.11 \mathrm{~s} \). By substitution, \( L=\left(\frac{T}{2 \pi}\right)^{2} g=\left(\frac{1.11 \mathrm{~s}}{2 \pi}\right)^{2} 9.81 \frac{\mathrm{m}}{\mathrm{s}^{2}}=0.307 \mathrm{~m}=30.7 \mathrm{~cm} \).