Question 262: Oscillations

Question

A particle performs linear S.H.M. At a particular instant, velocity of the particle is ' $u$ ' and acceleration is ' $\alpha$ ' while at another instant, velocity is ' $v$ ' and acceleration is ' $\beta$ ' $(0<\alpha<\beta)$. The distance between the two positions is

Accepted Answer

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

Answer

$\frac{u^2-v^2}{\alpha+\beta}$

Answer

$\frac{u^2+v^2}{\alpha+\beta}$

Answer

$\frac{u^2-v^2}{\alpha-\beta}$

Answer

$\frac{u^2+v^2}{\alpha-\beta}$

Oscillations

A particle performs linear S.H.M. At a particular instant, velocity of the particle is ' u ' and acceleration is ' \alpha ' while at another instant, velocity is ' v ' and acceleration is ' \beta ' (0<\alpha<\beta). The distance between the two positions is

Choose the correct answer: