Question

For a particle performing S.H.M. the equation $\frac {d^2x}{dt^2}$+αx=0. Then the time period of the motion will be

• A. 2πα
• B. $\frac {2π}{\sqrt {α}}$
• C. $\frac {2π}{α}$
• D. $2π\sqrt {α}$

Answer 1

Answer: (B)

$\frac {2π}{\sqrt {α}}$

Answer 2

In the given equation,
$\frac {d^2x}{dt^2}$ + αx = 0,
we can see that the equation represents simple harmonic motion (S.H.M.)
The general form of the equation for S.H.M. is:
$\frac {d^2x}{dt^2}$ + ω²x = 0,
where ω represents the angular frequency. Comparing this with the given equation, we can see that:
ω² = α.
The angular frequency (ω) of an oscillating system in S.H.M. is related to the time period (T) by the equation
ω = $\frac {2π}{T}$
Solving for T, we have:
T = $\frac {2π}{ω}$
T = $\frac {2π}{\sqrt {α}}$
Therefore, the correct answer is (B) $\frac {2π}{\sqrt {α}}$, which represents the time period of the motion.