A particle performs linear simple harmonic motion. If the displacement, acceleration and velocity of the particle are $x, a$ and $v$ respectively, which of the following graphs is /are correct

Question

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Answer 1

Let the displacement of particle performing SHM is given by, $x = A s i n w t$
Differentiating w.r.t time, $⟹ v = A w c o s w t$
Differentiating again w.r.t time, $⟹ a = - A w^{2} s i n w t$

Thus, $a = - w^{2} x$
Now, $v^{2} = A^{2} w^{2} (1 - s i n^{2} w t) ⟹ v^{2} = A^{2} w^{2} - \frac{1}{w ^{2}} a^{2}$
Thus $v^{2}$ vs $a^{2}$ should be a straight line with negative slope as represented by A.
Also, $v^{2} = A^{2} w^{2} (1 - s i n^{2} w t) ⟹ v^{2} = A^{2} w^{2} - w^{2} x^{2}$
Thus, $v^{2} + w^{2} x^{2} = A^{2} w^{2}$ which is an ellipse as represented by C.

Answer 2

Let the displacement of particle performing SHM is given by,

x = A s i n w t

Differentiating w.r.t time,

⟹ v = A w c o s w t

Differentiating again w.r.t time,

⟹ a = - A w^{2} s i n w t

Thus,

a = - w^{2} x

Now,

v^{2} = A^{2} w^{2} (1 - s i n^{2} w t) ⟹ v^{2} = A^{2} w^{2} - \frac{1}{w ^{2}} a^{2}

Thus

v^{2}

vs

a^{2}

should be a straight line with negative slope as represented by A.

Also,

v^{2} = A^{2} w^{2} (1 - s i n^{2} w t) ⟹ v^{2} = A^{2} w^{2} - w^{2} x^{2}

Thus,

v^{2} + w^{2} x^{2} = A^{2} w^{2}

which is an ellipse as represented by C.