Ncert Exercises & Solutions

The relations between acceleration and displacement of four particles are given below. Which one of the particles is executing simple harmonic motion?

1.	$a_1 = +2x$	2.	$a_1= +2x^2$
3.	$a_1= -2x^2$	4.	$a_1 = -2x$

Hint: Use the relation between the acceleration and displacement in SHM.

Explanation: In simple harmonic motion (SHM), the acceleration

$a$ is proportional to the negative displacement from the mean position. The general equation for SHM is:

$a = -\omega^2 x$
The acceleration should be linear with the displacement and negative sign, indicating that the acceleration is directed toward the mean position (restoring force).
let's analyse each option.
Option 1:

$a_1=+2x$
The acceleration is proportional to the displacement, but the sign is positive. This means the force is not directed toward the mean position, so this is not SHM.

Option 2: $a_2 = +2x^2$
The acceleration is proportional to $x^2,$ not to $x.$ This is non-linear and does not correspond to SHM.
Option 3: $a_3 = -2x^2$
Though the sign is negative, the acceleration is proportional to $x^2,$ which is non-linear. This is not SHM.
Option 4: $a_4 = -2x$
The acceleration is proportional to $x$ with a negative sign, which matches the form of SHM: $a = -\omega^2 x.$ This particle is undergoing SHM.
Hence, option (4) is the correct answer.

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