MHT CET · Physics · Oscillations

A particle is performing a linear simple harmonic motion of amplitude ‘A’. When it is midway between its mean and extreme position, the magnitudes of its velocity and acceleration are equal. What is the periodic time of the motion?

A $\frac{2 π}{\sqrt{3}} s$
B $\frac{\sqrt{3}}{2 π} s$
C $2 π \sqrt{3}$
D $\frac{1}{2 π \sqrt{3}} s$

Verified Solution

Answer & Solution

Correct Answer

(A) $\frac{2 π}{\sqrt{3}} s$

Step-by-step Solution

Detailed explanation

In linear simple harmonic motion, the velocity of particle is given by

v = ω \sqrt{A^{2} - x^{2}}

… (i)
where,

ω

= angular frequency
A = maximum displacement of amplitude
and x= displacement from mean position.
The acceleration of a particle in simple harmonic motion, (SHM) is given by

a = ω^{2} x

…. (ii)
Here,

x = \frac{A}{2}

Also,

v = a

(given)

ω \sqrt{(A^{2} - x^{2})} = ω^{2} x

[from Eqs. (i) and (ii), we get]

\Rightarrow \sqrt{(A^{2} - \frac{A^{2}}{4})} = ω \times \frac{A}{2} \Rightarrow \frac{\sqrt{3} A}{2} = ω \times \frac{A}{2}

\Rightarrow \frac{2 π}{T} = \sqrt{3} [∵ ω = \frac{2 π}{T}]

\Rightarrow T = \frac{2 π}{\sqrt{3}} s

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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