A particle performs simple harmonic motion (shm). What is the phase difference between the displacement and the acceleration of the particle?
The correct answer and explanation is:
The phase difference between the displacement and the acceleration of a particle performing simple harmonic motion is π radians or 180 degrees.
Explanation:
In simple harmonic motion (SHM), the displacement x(t)x(t) of the particle from its mean position can be described by a sinusoidal function such as: x(t)=Acos(ωt+ϕ)x(t) = A \cos(\omega t + \phi)
where:
- AA is the amplitude,
- ω\omega is the angular frequency,
- ϕ\phi is the phase constant,
- tt is time.
The velocity v(t)v(t) is the first derivative of displacement with respect to time: v(t)=dxdt=−Aωsin(ωt+ϕ)v(t) = \frac{dx}{dt} = -A \omega \sin(\omega t + \phi)
The acceleration a(t)a(t) is the second derivative of displacement: a(t)=d2xdt2=−Aω2cos(ωt+ϕ)a(t) = \frac{d^2x}{dt^2} = -A \omega^2 \cos(\omega t + \phi)
This shows acceleration is proportional to −cos(ωt+ϕ)-\cos(\omega t + \phi), which is the negative of the displacement function x(t)x(t).
Since acceleration and displacement are described by cosine functions with the same angular frequency and phase constant but acceleration has a negative sign, acceleration is exactly out of phase with displacement by π\pi radians or 180 degrees. When displacement is at its maximum positive value, acceleration is at its maximum negative value, pulling the particle back toward the mean position.
This phase difference reflects the restoring nature of the acceleration in SHM. The acceleration always acts in the opposite direction to displacement, trying to bring the particle back to equilibrium. Because of this relationship, displacement and acceleration are perfectly out of sync by half a cycle.
In summary, the phase difference between displacement and acceleration in SHM is π\pi radians, showing they are opposite in phase.