Question

The amplitude of a particle executing $SHM$ is $4 cm$. At the mean position, the speed of the particle is $16 cm / s$. The distance of the particle from the mean position at which the speed of the particle becomes $8 \sqrt{3} cm / sec$ will be

• A. $2 \sqrt{3}$
• B. $\sqrt{3}$
• C. 1
• D. 2

Answer 1

Answer: (D)

$\omega=\frac{V_{\max }}{a} \Rightarrow \frac{16}{4}=4$
$V=\omega \sqrt{a^{2}-y^{2}}$
$\Rightarrow 8 \sqrt{3}=4 \sqrt{16-y^{2}}$
$\Rightarrow 2 \sqrt{3}=\sqrt{16-y^{2}}$
$\Rightarrow 12=16- y ^{2}$
$\Rightarrow y ^{2}=4 \quad$ (squaring both side)
$y=2 cm$