Question

For a particle performing S.H.M. the equation $ \frac{d^2x}{dt^2} + \alpha x = 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

The Correct Option is B

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.