Question 249: Oscillations

Question

A particle executes a linear S.H.M. In two of its positions the velocities are $\mathrm{V}_1, \mathrm{~V}_2$ and accelerations are $\mathrm{a}_1$ and $\mathrm{a}_2$ respectively $\left(0

Accepted Answer

When velocity is $V_1$ and acceleration is $a_1$, let the position of particle be $\mathrm{x}_1$.
When velocity is $\mathrm{V}_2$ and acceleration is $\mathrm{a}_2$, let the position of particle be $\mathrm{x}_2$.
If $\omega$ is the angular frequency then,
$$ a_1=\omega^2 x_1 $$
and $\mathrm{a}_2=\omega^2 \mathrm{x}_2$.
$$ 	herefore \quad \mathrm{a}_1+\mathrm{a}_2=\omega^2\left(\mathrm{x}_1+\mathrm{x}_2ight) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$$
Also, velocity of particle at particular instant can be given as,
$$ V_1^2=\omega^2 A^2-\omega^2 x_1^2 $$
and $V_2{ }^2=\omega^2 A^2-\omega^2 x_2^2$
i.e., $V_2{ }^2-V_1{ }^2=\omega^2\left(x_1^2-x_2^2ight)$
$$ V_2^2-V_1^2=\omega^2\left(x_1-x_2ight)\left(x_1+x_2ight) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$$
from equation (i) we get
$$ \begin{aligned} & V_2^2-V_1^2=\left(x_1-x_2ight)\left(a_1+a_2ight) \ 	herefore \quad & x_1-x_2=\frac{V_2^2-V_1^2}{a_1+a_2} \ & 	ext { or } x_2-x_1=\frac{V_1^2-V_2^2}{a_1+a_2} \end{aligned} $$

Answer

$\frac{V_1^2-V_2^2}{a_1-a_2}$

Answer

$\frac{V_2^2-V_1^2}{a_1-a_2}$

Answer

$\frac{V_1^2-V_2^2}{a_1+a_2}$

Answer

$\frac{v_2^2-v_1^2}{\left(a_1^2+a_2^2\right)}$

Oscillations

A particle executes a linear S.H.M. In two of its positions the velocities are \mathrm{V}_1, \mathrm{~V}_2 and accelerations are \mathrm{a}_1 and \mathrm{a}_2 respectively $\left(0

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