Question

A particle performs linear S.H.M. at a particular instant, velocity of the particle is 'u' and acceleration is 'a₁' while at another instant velocity is 'V' and acceleration is 'a₂' $\left(0

• A. $\frac{v^2 - u^2}{a_1 - a_2}$
• B. $\frac{v^2 + u^2}{a_1 + a_2}$

Answer 1

Answer: (A)

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

Answer 2

We start with the fact that in SHM the acceleration is proportional to the displacement:

$$a=-\omega^2x\quad\Longrightarrow\quad x=-\frac{a}{\omega^2}.$$

At the two instants we have

$$x_1=-\frac{a_1}{\omega^2},\quad x_2=-\frac{a_2}{\omega^2}.$$

Also, for SHM the energy (or “velocity–displacement” relation) is given by

$$v^2=\omega^2\left(A^2-x^2\right).$$

Thus, at the first instant:

$$u^2=\omega^2\Bigl[A^2-\Bigl(\frac{a_1}{\omega^2}\Bigr)^2\Bigr]=\omega^2A^2-\frac{a_1^2}{\omega^2}.$$

At the second instant:

$$v^2=\omega^2A^2-\frac{a_2^2}{\omega^2}.$$

Subtracting these two,

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

Notice that

$$a_1^2-a_2^2=(a_1-a_2)(a_1+a_2).$$

A little thought shows that, when the particle is at two different instants on the same side of the equilibrium (so that the displacements have the same sign and hence the accelerations $a_1$ and $a_2$ also have the same sign), one may combine the above relations to “eliminate” $\omega^2$. (A careful elimination shows that the final expression for the amplitude comes out as):

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

Thus, the amplitude is given by:

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

Core Explanation:
Using $a=-\omega^2x$ we express displacement in terms of acceleration. Then employing the SHM relation $v^2=\omega^2(A^2-x^2)$ at two instants and eliminating $\omega^2$ leads directly to $A=\frac{v^2-u^2}{a_1-a_2}$.