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Simple Harmonic Motion question

2005 · Shift 0 · Q157

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Simple Harmonic Motion
2005 · Shift 0 · Q157

Simple Harmonic Motion question

2005 · Shift 0 · Q157

JEE MainPhysicsSimple Harmonic MotionMCQ+4 / −1

If a simple harmonic motion is represented by

{{{d^2}x} \over {d{t^2}}} + \alpha x = 0.

its time period is

A
${{2\pi } \over {\sqrt \alpha }}$
B
${{2\pi } \over \alpha }$
C
$2\pi \sqrt \alpha$
D
$2\pi \alpha$

View written solutionFree

Correct answer: A

The standard differential equation of simple harmonic motion is

$\frac{d^2x}{dt^2} + \omega^2 x = 0$

where $\omega$ is the angular frequency.

The given equation is

$\frac{d^2x}{dt^2} + \alpha x = 0$

Comparing with the standard form,

$\omega^2 = \alpha$

So,

$\omega = \sqrt{\alpha}$

The time period of SHM is

$T = \frac{2\pi}{\omega}$

Substituting $\omega = \sqrt{\alpha}$ ,

$T = \frac{2\pi}{\sqrt{\alpha}}$

Now compare with the options:

A: $\dfrac{2\pi}{\sqrt{\alpha}}$ ✅
B: $\dfrac{2\pi}{\alpha}$
C: $2\pi\sqrt{\alpha}$
D: $2\pi\alpha$

Therefore, the correct option is A.

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