Here are the detailed notes on deriving the equations of force in SHM and the time period in terms of kk and mm, with math formatted in unrendered LaTeX:
1. Force in SHM
The force acting on a particle in SHM is governed by Hooke’s Law. It is directly proportional to the displacement and acts in the opposite direction.
Equation of Force:
From Newton’s second law, the force is:
[math]F = m a[/math]
Using the equation for acceleration in SHM:
[math]a(t) = -\omega^2 x(t)[/math],
we substitute:
[math]F = m \cdot (-\omega^2 x)[/math]
Thus, the force equation becomes:
[math]F = -m \omega^2 x[/math]
Since the restoring force in SHM is also given by:
[math]F = -k x[/math]
Equating the two expressions for force:
[math]-m \omega^2 x = -k x[/math]
Canceling [math]x[/math] (as [math]x \neq 0[/math]):
[math]m \omega^2 = k[/math]
This relates the angular frequency [math]\omega[/math] to the mass [math]m[/math] and spring constant [math]k[/math]:
[math]\omega = \sqrt{\frac{k}{m}}[/math]
2. Time Period in Terms of k and m
The time period [math]T[/math] is the time taken for one complete oscillation. It is related to angular frequency [math]\omega[/math] by:
[math]\omega = \frac{2\pi}{T}[/math]
From the above relation for [math]\omega[/math]:
[math]\omega = \sqrt{\frac{k}{m}}[/math]
Substituting:
[math]\frac{2\pi}{T} = \sqrt{\frac{k}{m}}[/math]
Rearranging for [math]T[/math]:
[math]T = 2\pi \sqrt{\frac{m}{k}}[/math]
Final Equations:
- Force:
[math]F = -k x[/math]
[math]F = -m \omega^2 x[/math]
Angular Frequency:
[math]\omega = \sqrt{\frac{k}{m}}[/math]
Time Period:
[math]T = 2\pi \sqrt{\frac{m}{k}}[/math]
Let me know if you need further clarifications or additional derivations!
