We may solve the above differential equation for the SHM of a particle. Let us assume that the particle, executing SHM is oscillating, (about its mean position, O) between the points A and B; then |OA| = |OB| = a, is known as the ‘amplitude’ of the SHM.
The differential equation, for a SHM, can be written as
where v = dx/dt , is the instantaneous velocity of the particle.
The value of the constant can be found by remembering that the instantaneous velocity, of the particle, becomes zero when it is at either of its extreme positions (A or B). Thus
The instantaneous displacement, of a particle, executing a SHM, is thus a sinusoidal function of time.
The form of this sinusoidal function depends on the choice of the ‘zero’ of time. If, at t = 0 (i.e., at the start), the particle is at its mean position, we have x = 0 at t = 0.
∴ In this case,
0 = a sin (0 + ϕ) = a sinϕ
This implies that ϕ = 0 . Thus if the ‘zero’ of time is chosen to be the (time) instant at which the particle is in its mean position, the instantaneous displacement, x, of the particle, is a SINE function of time
x = a sin ωt
If, however, the ‘zero’ of the time is taken to be the instant at which the particle is at its extreme position
Thus, if the ‘zero’ of time is taken to be instant at which the particle is at either of the extreme positions, its instantaneous displacement is a COSINE function of time.
When the ‘zero’ of time is taken as the instant at which the particle is somewhere in between its mean and (either) extreme position, the instantaneous displacement will be described by the sinusoidal function.
x = a sin (ωt + ϕ)
The value of ϕ will be determined by the value of x (0 < |x| < a) at t = 0.
