Question

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

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

Answer 1

Answer: (D)

2 cm

Answer 2

At mean position velocity is maximum

i.e., $v_{\max} = \omega a \Rightarrow \omega = \frac{v_{\max}}{a} = \frac{16}{4} = 4$

∴ $v = \omega\sqrt{a^{2} - y^{2}} \Rightarrow 8\sqrt{3} = 4\sqrt{4^{2} - y^{2}}$

$\Rightarrow 192 = 16(16 - y^{2}) \Rightarrow 12 = 16 - y^{2} \Rightarrow y = 2cm.$.