Question

A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position the velocity of the particle is 10 cm/s. The distance of the particle from the mean position when its speed becomes 5 cm/s is

• A. $\sqrt{3}cm$
• B. $\sqrt{5}cm$
• C. $2(\sqrt{3})cm$
• D. $2(\sqrt{5})cm$

Answer 1

Answer: (C)

$2(\sqrt{3})cm$

Answer 2

$v_{\max}$ ⇒ $\omega = \frac{v_{\max}}{a\frac{10}{4}}$

Now, $v = \omega\sqrt{a^{2} - y^{2}}$⇒ $v^{2} = \omega^{2}(a^{2} - y^{2})$ ⇒ $y^{2} = a^{2} - \frac{v^{2}}{\omega^{2}}$

⇒ $T \propto \sqrt{m} = 2\sqrt{3}$cm