A) When v is maximum, a is minimum
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
Correct Answer: A
Solution :
In simple harmonic motion, the displacement equation is,\[y=A\sin \omega t\] Where A is amplitude of the motion. Velocity, \[v=\frac{dy}{dt}=A\,\omega \cos \omega t\] \[v=A\omega \sqrt{1-{{\sin }^{2}}\omega t}\] \[v=\omega \sqrt{{{A}^{2}}-{{y}^{2}}}\] (i) Acceleration, \[a=\frac{dv}{dt}=\frac{d}{dt}(A\omega cos\omega t)\] \[a=-A{{\omega }^{2}}\sin \omega t\] \[a=-A{{\omega }^{2}}\sin \omega t\] \[a=-{{\omega }^{2}}y\] (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 (i.e., zero) or vice versa.
You need to login to perform this action.
You will be redirected in
3 sec
