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{d y}{d t}=A\, \omega \,\cos\, \omega t $
$v =A\, \omega \sqrt{1-\sin ^{2} \omega t}$
$\nu =\omega \sqrt{A^{2}-y^{2}}\ldots$(i)
Acceleration, $a=\frac{d v}{d t}=\frac{d}{d t}(A \omega\, \cos\, \omega t)$
$ a=-A \omega^{2} \sin\, \omega t $
$ a =-\omega^{2} y \ldots$(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.