Question

Which of the following statements are true, for a particle performing S.H.M with amplitude $A$ and angular frequency $\omega$?

• A. $v_{max}=\pm \omega A,$ at mean position
• B. $a_{max}=\pm {\omega }^{2}A$, at extreme position
• C. $a_{min}=0,$ at mean position
• D. $v_{min}=0$, at extreme position

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $v_{max}=\pm \omega A,$ at mean position
B $a_{max}=\pm {\omega }^{2}A$, at extreme position
C $a_{min}=0,$ at mean position
D $v_{min}=0$, at extreme position

Velocity-Displacement relationship
$v=\pm \omega \sqrt{A^2-x^2}$
$\Rightarrow v_{max}=\pm \omega A\text{at}x=0\to$ Mean position.
$\&v_{min}=0\text{at}x=A\to$ Extreme position.

Acceleration-Displacement relationship
$a=-{\omega }^{2}x$
$\Rightarrow |a_{max}|={\omega }^{2}A\text{at}x=\pm A\to$ extreme position.
$\&|a_{min}|=0\text{at}x=0\to$ Mean position.
Hence, options (a), (b), (c) and (d) are correct.