Question

Which one of the following statements is true for the speed $v$ and the acceleration $a$ of a particle executing simple harmonic motion?

• A. When v is maximum, a is maximum
• B. Value of a is zero, whatever may be the, value of v
• C. When v is zero, a is zero
• D. When v is maximum, a is zero

Answer 1

Answer: (D)

In simple harmonic motion, the displacement equation is, $y=A\sin \omega t$
where $A$ is amplitude of the motion.
Velocity, $v=\frac{dy}{dt}=A\omega \cos \omega t$
$v=A\omega \sqrt{1-{\sin }^2\omega t}$
$\nu =\omega \sqrt{A^2-y^2}\dots$(i)
Acceleration, $a=\frac{dv}{dt}=\frac{d}{dt}(A\omega \cos \omega t)$
$a=-A{\omega }^2\sin \omega t$
$a=-{\omega }^{2}y\dots$(ii)
When $y=0;v=A\omega =v_{\max }$
$a=0=a_{\min }$
When $y=A;v=0=v_{\min }$
$a=-{\omega }^{2}A=a_{\max }$
Hence, it is clear that when $v$ is maximum, then $a$ is minimum (ie, zero) or vice-versa.