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
Correct Answer: D
Solution :
| In simple harmonic motion, the displacement equation is, \[y=A\,\sin \,\,\omega t\] where A is amplitude of the motion. |
| Velocity, \[v=\frac{dy}{dt}=A\,\,\omega \,\,\cos \,\,\omega t\] |
| \[y=A\,\,\omega \sqrt{1-{{\sin }^{2}}\omega t}\] |
| \[y=\,\,\omega \sqrt{{{A}^{2}}-{{y}^{2}}}\] ...(i) |
| Acceleration, \[a=\frac{dy}{dt}=\frac{d}{dt}(A\omega \,\cos \,\omega t)\] |
| \[a=-A{{\omega }^{2}}\,\,\sin \,\,\omega t\] |
| \[a=-{{\omega }^{2}}\,\,y\] ...(ii) |
| When \[y=0;\,\,v=A\omega ={{y}_{\max }}\] |
| \[a=0\,\,={{a}_{\min }}\] |
| 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. |
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