question_answer

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

A)  0.5\[{{T}_{i}}=2{{T}_{n}}-T\] 

B)                                        \[{{0}^{o}}C\]                  

C)        0.707\[{{T}_{n}}=\frac{{{T}_{i}}}{2}\]                     

D)        zero

Correct Answer: A

Solution :

The displacement equation of particle executing SHM is \[^{40}Ar\]                   ...(i) Velocity, \[^{40}k\]   ...(ii) Acceleration, \[2.95\times {{10}^{11}}yr\]                        ?(iii) Fig. (i) is a plot of Eq. (i) with \[\phi =0\]. Fig. (ii) shows Eq. (ii) also with \[\phi =0\]. 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 \[2.95\times {{10}^{9}}yr\]. Similarly, the acceleration curve of A is shifted (to the left) by \[4.37\times {{10}^{9}}yr\] relative to the velocity curve of v. This implies that velocity is \[4.37\times {{10}^{11}}yr\] \[\sigma \] out of phase with the displacement and the acceleration is \[\rho \] \[\lambda \] out of phase with the velocity but \[\frac{\sigma }{\rho \lambda }\] out of phase with displacement.