If the frequency of the motion of a simple harmonic oscillator is doubled, by what factor does the maximum speed of the oscillator change?
Physics · High School · Thu Feb 04 2021
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Answer: The maximum speed \(v_{max}\) of a simple harmonic oscillator (SHO) is given by the product of the angular frequency \(\omega\) and the amplitude \(A\):
\[v_{max} = \omega A\]
Where: - \(v_{max}\) is the maximum speed, - \(\omega\) is the angular frequency, and - \(A\) is the amplitude of the motion.
The angular frequency \(\omega\) is related to the frequency \(f\) of the oscillator by the formula:
\[\omega = 2\pi f\]
So, if the frequency \(f\) is doubled, the angular frequency \(\omega\) is also doubled, assuming that the amplitude \(A\) remains constant.
Hence, if the frequency is doubled, the new maximum speed \(v'_{max}\) will be:
\[v'_{max} = (2\omega) A = 2(\omega A) = 2v_{max}\]
Therefore, the maximum speed of the oscillator is also doubled.