Answer:
30720 oscillation.
Explanation:
Frequency: This can be defined as the number of oscillation completed in a seconds. The S.I unit of frequency is Hertz.
From the question,
The frequency of the string set is 512 Hz,
These means that the spring set complete 512 oscillation in one seconds
I.e
if its completes in one seconds 512 oscillation.
Then, in one minutes, it will complete x oscillation.
x = 512×1min/1seconds
But 1 minutes = 60 seconds
x = 512×60/1
x = 30720 oscillation.
The number of oscillations made by guitar string in each minute will be 30720.
Given data:
The frequency of vibration is, f = 512 Hz.
The given problem is based on the concept of frequency of vibration. This can be defined as the number of oscillation completed in a seconds.
The number of oscillation in 1 second is,
[tex]f = \dfrac{n}{t} \\\\512 = \dfrac{n}{ 1}\\\\n = 512 \;\rm oscillations[/tex]
Since, 1 minute = 60 seconds.
So, oscillations in 60 seconds is,
[tex]n = 512 \times 60\\n = 30720 \;\rm oscillations[/tex]
Thus, we can conclude that the number of oscillations in each minute will be 30720.
Learn more about the frequency of oscillation here:
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