Question

The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is

• A. $0.5Ï€$
• B. $Ï€$
• C. $0.707Ï€$
• D. zero

Answer 1

Answer: (A)

The displacement equation of particle executing $SHM$ is
$x=a\cos (ωt+ϕ)$(i)
Velocity, $v=\frac{dx}{dt}=−aω\sin (ωt+ϕ)$ (ii)
Acceleration,

$A=\frac{dv}{dt}=−a{ω}^2\cos (ωt+ϕ)$ (iii)
Fig. (i) is a plot of Eq. (i) with $ϕ=0$. Fig. (ii) shows Eq. (ii) also with $ϕ=0$. Fig. (iii) is a plot of Eq. (iii). It should be noted that in the figures the curve of $v$ is shifted (to the left) from the curve of $x$ by one-quarter period $\left(\frac{1}{4}T\right)$. Similarly, the acceleration curve of $A$ is shifted (to the left) by $\frac{1}{4}T$ relative to thevelocity curve of $v$. This implies that velocity is $90^{∘}(0.5π)$ out of phase with the displacement and the acceleration is $90^{∘}(0.5π)$ out of phase with the velocity but $180^{∘}(π)$ out of phase with displacement.