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The frequency of oscillation for a mass \[m\] will be proportional to ___

Last updated date: 13th Jun 2024
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Hint: We will briefly look at the force balance equation for a harmonic oscillator to determine the different forces that act on the mass. Then we will assign the value of frequency of oscillation and look at its dependence on mass.

Complete step by step answer:
When a mass is tied to spring, it will execute a simple harmonic motion. The frequency of oscillation of this mass depends on the mass as well as the spring constant of the spring. Let the mass of the block be $m$ and the spring constant be $k$.
Then the only force acting on the block will be the restoring force of the spring which can be calculated from Hooke’s law as
$F = - kx$
From Newton’s second law, we know that this force will cause a net acceleration on the block which can be written mathematically as
$F = ma$
$ \Rightarrow - kx = ma$
Since we know that $a = \dfrac{{{d^2}x}}{{d{t^2}}}$, we can write
$m\dfrac{{{d^2}x}}{{d{t^2}}} + kx = 0$
Dividing both sides by $m$, we get
 \[\dfrac{{{d^2}x}}{{d{t^2}}} + \dfrac{k}{m}x = 0\]
The term $k/m$ in the above equation is replaced by ${\omega ^2} = \dfrac{k}{m}$ where $\omega $ is the angular frequency of the oscillation of the mass.
Then the value of $\omega $ will be
$\omega = \sqrt {\dfrac{k}{m}} $
Since the angular frequency is related to the frequency as $\omega = 2\pi f$, we can write
$f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} $
Hence we can see that the frequency of oscillation is proportional to
$f \propto \sqrt {\dfrac{k}{m}} $

Note: We should be aware of the force balance equation for a harmonic oscillator and its general form. While the angular frequency and the frequency of oscillation have the same dependence on the spring constant and mass, they’re different quantities and must not be confused.