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in Physics by (75.2k points)
The equation of a particle executing `SHM` is `(d^(2)x)/(dt^(2))=-omega^(2)x`. Where `omega=(2pi)/("time period")`. The velocity of particle is maximum when it passes through mean position and its accleration is maximum at extremeposition. The displacement of particle is given by `x=A sin(omegat+theta)` where `theta`-initial phase of motion. `A`-Amplitude of motion and T-Time period
The average velocity during motion of particle from one extreme point:
A. `(2v)/(pi)`
B. `(2A)/(l)`
C. `(3A)/(t)`
D. `(v)/(pi)`

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1 Answer

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by (67.7k points)
Correct Answer - A

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