A particle executes simple harmonic motion with an amplitude of 4… — Solution | JEE Physics

NVAEasyJEE 2024Simple Harmonic Motion (SHM)

A particle executes simple harmonic motion with an amplitude of $4 \, \text{cm}$ . At the mean position, the velocity of the particle is $10 \, \text{cm/s}$ . The distance of the particle from the mean position when its speed becomes $5 \, \text{cm/s}$ is $\sqrt{\alpha}$ cm, where $\alpha =$ ?

Answer

Correct answer:12

Step-by-step solution

Standard Method

Given: Amplitude $A = 4 \, \text{cm}$ , maximum speed at mean position $v_{\max} = 10 \, \text{cm/s}$ , and speed at displacement $x$ is $5 \, \text{cm/s}$ .

Find: The value of $\alpha$ if $x = \sqrt{\alpha} \, \text{cm}$ .

For simple harmonic motion, at the mean position:

v_{\max} = A\omega = 10

At displacement $x$ :

v = \omega\sqrt{A^2 - x^2} = 5

Dividing the two relations:

\frac{A}{\sqrt{A^2 - x^2}} = 2

So,

A^2 = 4(A^2 - x^2)

3A^2 = 4x^2

Hence,

x = \frac{A\sqrt{3}}{2}

Using $A = 4 \, \text{cm}$ ,

x = \frac{4\sqrt{3}}{2} = 2\sqrt{3} \, \text{cm}

Since $x = \sqrt{\alpha} \, \text{cm}$ ,

\sqrt{\alpha} = 2\sqrt{3} = \sqrt{12}

Therefore, $\alpha = 12$ .

The required numerical value is $12$ .

Common mistakes

Using the linear SHM relation incorrectly by assuming speed is directly proportional to displacement. In SHM, speed depends on $\sqrt{A^2 - x^2}$ , not on $x$ itself. Use the standard speed formula before substituting values.
Substituting $A = 4$ too early without first forming the ratio of the two equations. The ratio cancels $\omega$ cleanly and makes the algebra simpler. First divide the equations, then solve for $x$ .
Confusing $x = 2\sqrt{3}$ with $\alpha$ directly. Since the question gives $x = \sqrt{\alpha}$ cm, you must square the value of $x$ to obtain $\alpha$ . Therefore, $\alpha = 12$ , not $2\sqrt{3}$ .

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