Question

The differential equation representing the SHM of a particuleis $\frac{9d^{2}y}{d{t}^2}+4y=0$. The time period of the particle is given by

• A. $\frac{\pi }{3}sec$
• B. $\pi sec$
• C. $\frac{2}{\pi }3sec$
• D. $3\pi sec$

Answer 1

The correct option is D $3\pi sec$

Given,

$9\frac{d^{2}y}{d{t}^2}+4y=0$. . . .. . . . . .(1)

acceleration, $a=\frac{d^{2}y}{d{t}^2}$

equation (1) becomes,

$9a+4y=0$

$a=-\frac{4}{9}y$ ($a=-{\omega }^{2}y$)

${\omega }^2=\frac{4}{9}$

$\omega =\frac{2}{3}$

$\frac{2\pi }{T}=\frac{2}{3}$

$T=3\pi sec$

The correct option is D.