Question 274: Oscillations

Question

A particle performs linear S.H.M. at a particular instant, velocity of the particle is ' $u$ ' and acceleration is ' $\mathrm{a}_1$ ' while at another instant velocity is ' V ' and acceleration is ' $a_2$ ' $\left(0

Accepted Answer

When velocity is $u$ and acceleration is $a_1$, let the position of particle be $\mathrm{x}_1$.
When velocity is $v$ and acceleration is $a_2$, let the position of particle be $x_2$.
If $\omega$ is the angular frequency then, $a_1=\omega^2 x_1$
and $a_2=\omega^2 x_2$
$	herefore \quad \mathrm{a}_1+\mathrm{a}_2=\omega^2\left(\mathrm{x}_1+\mathrm{x}_2ight)\quad	ext{... (i)}$
Also, velocity of particle at particular instant can be given as,
$$\begin{aligned}
& u^2=\omega^2 A^2-\omega^2 x_1^2 \
& 	ext { and } v^2=\omega^2 A^2-\omega^2 x_2^2 \
& 	ext { i.e., } v^2-u^2=\omega^2\left(x_1^2-x_2^2ight) \
& v^2-u^2=\omega^2\left(x_1-x_2ight)\left(x_1+x_2ight)\quad	ext{.... (ii)}
\end{aligned}$$
from equation (i) we get
$$\begin{aligned}
	herefore \quad & x_1-u_2= \frac{v^2-u^2}{a_1+a_2} \
& 	ext { or } x_2-x_1=\frac{u^2-v^2}{a_1+a_2}
\end{aligned}$$

Answer

$\frac{v^2-u^2}{a_1-a_2}$

Answer

$\frac{v^2+u^2}{a_1+a_2}$

Answer

$\frac{u^2+v^2}{a_1-a_2}$

Answer

$\frac{u^2-V^2}{a_1+a_2}$

Oscillations

A particle performs linear S.H.M. at a particular instant, velocity of the particle is ' u ' and acceleration is ' \mathrm{a}_1 ' while at another instant velocity is ' V ' and acceleration is ' a_2 ' $\left(0

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