Question
Which one of the following statements is true for the speed $$v$$ and the acceleration $$\alpha $$ of a particle executing simple harmonic motion?
A.
When $$v$$ is maximum, $$\alpha $$ is maximum
B.
Value of $$\alpha $$ is zero, whatever may be the value of $$v$$
C.
When $$v$$ is zero, $$\alpha $$ is Zero
D.
When $$v$$ is maximum, $$\alpha $$ is zero
Answer :
When $$v$$ is maximum, $$\alpha $$ is zero
Solution :
In simple harmonic motion, the displacement equation is, $$x = a\sin \omega t$$
where, $$a$$ is the amplitude of the motion.
$$\eqalign{
& {\text{Velocity, }}v = \frac{{dx}}{{dt}} = a\omega \cos \omega t \cr
& v = a\omega \sqrt {1 - {{\sin }^2}\omega t} \cr
& v = \omega \sqrt {{a^2} - {x^2}} \,......\left( {\text{i}} \right) \cr
& {\text{Acceleration, }}\alpha = \frac{{dv}}{{dt}} = \frac{d}{{dt}}\left( {a\omega \cos \omega t} \right) \cr
& \alpha = - a{\omega ^2}\sin \omega t \cr
& \alpha = - {\omega ^2}x\,......\left( {{\text{ii}}} \right) \cr
& {\text{When}}\,x = 0,v = a\omega = {v_{\max }} \cr
& \alpha = 0 = {\alpha _{\min }} \cr
& {\text{When}}\,x = a,v = 0 = {v_{\min }} \cr
& \alpha = - {\omega ^2}a = {\alpha _{\max }} \cr} $$
Hence, it is clear that when $$v$$ is maximum, then $$\alpha $$ is minimum (i.e. zero) or vice-versa.