A) a straight line
B) a parabola
C) a hyperbola
D) an ellipse
Correct Answer: D
Solution :
The velocity (u) of a particle in SHM changes with displacement y as follows \[u=\omega \sqrt{{{a}^{2}}-{{y}^{2}}}\] ? (i) where a is amplitude and (0 is angular speed. Squaring both sides of Eq. (i), we get \[{{u}^{2}}={{\omega }^{2}}\,({{a}^{2}}-{{y}^{2}})\] \[\frac{{{u}^{2}}}{{{\omega }^{2}}}={{a}^{2}}-{{y}^{2}}\] \[\Rightarrow \frac{{{u}^{2}}}{{{\omega }^{2}}\,\,{{a}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}}=1\] which is general equation of an ellipse. Note: When displacement is maximum, then velocity is zero.
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