derive an expression for particle performing SHM
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Acceleration of S.H.M. = d^2x/dt^2 = – ω^2x
Explanation:
As we know that:
The acceleration is velocity per unit time. its unit is m/s^2.
The acceleration of a particle performing S.H.M can be calculated by using the following relation.
S.H.M. is d2x/dt2 + (k/m)x = 0
This is the differential equation of linear simple harmonic motion.
- Where d2x/dt2 is the acceleration of the particle.
- x is the displacement of the particle.
- m is the mass of the particle
- k is the force constant.
- We know that k/m = ω2 where ω is the angular frequency.
Therefore, d^2x/dt^2 +ω^2 x = 0
Acceleration of S.H.M. = d^2x/dt^2 = – ω^2x
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