Question

The reading on a metronome indicates the number of oscillations per minute. What are the frequency and period of the metronome's vibration when the metronome is set at 180?

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Given data: n=180n = 180

In order to solve this problem we have to assume that, in the case of metronome, frequency is given as number of oscillations in one minute. Therefore we will have:

f=n1minf = \dfrac{n}{1\, \mathrm{min}}

When we express it into seconds we get (1min=60s1\, \mathrm{min} = 60\, \mathrm{s}):

f=n60sf = \dfrac{n}{60\, \mathrm{s}}

And we can calculate period later as:

T=1fT = \dfrac{1}{f}

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