Question

A metronome uses an inverted pendulum rod to control tempo. A weight can be slid up the pendulum rod to decrease tempo, or down the pendulum rod to increase tempo. A second fixed weight is hidden inside the metronome at the base of the pendulum. The longer the distance between the two weights, the slower the tempo. This equation is used to calculate the frequency of the metronome pendulum:
$f = \frac{ \sqrt{ \frac{g}{l} } }{2\pi}$
Each time the metronome's pendulum completes one back and forth cycle, it produces two ticks or beats. To calibrate the metronome for beats per minute (bpm) the manufacturer needs to know the length of the pendulum for various tempos. Solve
$f = \frac{ \sqrt{ \frac{g}{l} } }{2\pi}$
for L.​

Answer 1

Answer:

Part 1

Multiply both sides by 2π

2πf = √(g/L)

Square both sides

4π²f²= g/L

Invert both sides

1/(4π²f²) = L/g

Multiply both side by g

g/(4π²f²) = L

We usually write an equation with the subject (L) n the left

L = g/(4π²f²)

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Part 2

Using the above equation with the given values:

L = g/(4π²f²)

. .= 9.8 / (4π² x 1.6²)

. .= 0.097m (= 9.7cm)

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By the way, where it says

“f=1.6 then there are 2 beats a second, or 192 beats per minute (bpm).”

this should say

“f=1.6 then there are 3.2 beats a second, or 192 beats per minute (bpm)”

Step-by-step explanation: