Q. The amplitude of a particle executing SHM is 4 cm. At the mean position the speed of the particle is 16$cms^{- 1}$ . The distance of the particle from the mean position at which the speed of the particle become $8\sqrt{3}\ cms^{- 1}$, will be
A) $2\sqrt{3}$ cm
B) $\sqrt{3}$ cm
C) 1 cm
D) 2 cm
Explanation
(d) At the mean position, the speed will be maximum
$v_{\max} = 16\ cms^{- 1} = a\omega = 4\omega$,
So, $\omega = \frac{16}{4} = 4\ \frac{rad}{s}$
and $v = \omega\sqrt{A^{2} - y^{2}}$
$8\sqrt{3} = 4\sqrt{A^{2} - y^{2}}$
or $A^{2} - y^{2} = 12$
or $y^{2} = A^{2} - 12 = (4)^{2} - 12 = 4$
or $y = 2$ cm
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.