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Question

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

The frequency of oscillation for a mass mm will be proportional to ___

Explanation

Solution

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 mm and the spring constant be kk.
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=kxF = - 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=maF = ma
kx=ma\Rightarrow - kx = ma
Since we know that a=d2xdt2a = \dfrac{{{d^2}x}}{{d{t^2}}}, we can write
md2xdt2+kx=0m\dfrac{{{d^2}x}}{{d{t^2}}} + kx = 0
Dividing both sides by mm, we get
d2xdt2+kmx=0\dfrac{{{d^2}x}}{{d{t^2}}} + \dfrac{k}{m}x = 0
The term k/mk/m in the above equation is replaced by ω2=km{\omega ^2} = \dfrac{k}{m} where ω\omega is the angular frequency of the oscillation of the mass.
Then the value of ω\omega will be
ω=km\omega = \sqrt {\dfrac{k}{m}}
Since the angular frequency is related to the frequency as ω=2πf\omega = 2\pi f, we can write
f=12πkmf = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}}
Hence we can see that the frequency of oscillation is proportional to
fkmf \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.