Question

A particle starts S.H.M. from the mean position. Its amplitude is A and time period is T. At the time when its speed is half of the maximum speed. Its displacement y is

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

Answer 1

Answer: (C)

$A\sqrt{3}/2$

Answer 2

$v = \omega\sqrt{a^{2} - y^{2}}$ ⇒ $\frac{a\omega}{2} = \omega\sqrt{a^{2} - y^{2}}$⇒ $\frac{a^{2}}{4} = a^{2} - y^{2}$

⇒ $y = \frac{\sqrt{3}A}{2}$ [As $v = \frac{v_{\max}}{2\frac{a\omega}{2}}$