Question

The differential equation for a particle of mass 2kg executing SHM is $2\frac{d^{2}x}{d{t}^2}+x=0$. If amplitude of oscillation is 3 cm, then maximum velocity of the particle will be

• A. $3\sqrt{0.5}$ cm/s
• B. $\sqrt{0.5}$ cm/s
• C. $\sqrt{0.5}/3$ cm/s
• D. $2\sqrt{0.5}$ cm/s

Answer 1

The correct option is A $3\sqrt{0.5}$ cm/s

Comparing the equation $2\frac{d^{2}x}{d{t}^2}+x=0$ with the standard equation $m\frac{d^{2}x}{d{t}^2}+kx=0$, we get, k=1 and $\omega =\sqrt{(}k/m)=\sqrt{(}0.5)$
The maximum velocity of the particle is $A\omega =3\sqrt{(}0.5)$ cm/s
The correct option is (a)