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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