Question

The amplitude of a particle executing $SHM$ is $4cm$. At the mean position the speed of the particle is $16cm/sec$. 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. $2\sqrt{3}cm/s$
• B. $\sqrt{3}cm/s$
• C. $1cm$
• D. $2cm$

Answer 1

The correct option is D $2cm$

At mean position velocity is maximum

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

The velocity can also be represented as:

$\therefore v=\omega \sqrt{a^2-y^2}$

$\Rightarrow 8\sqrt{3}=4\sqrt{4^2-y^2}$

$\Rightarrow 192=16(16-y^2)\Rightarrow 12=16-y^2$

$\Rightarrow y=2cm$.