Question

A particle executes simple harmonic motion with an amplitude of 4 cm. At the

mean position the velocity of the particle is 10 cm/s. The distance of the

particle from the mean position when its speed becomes 5 cm/s is

[EAMCET (Med.) 2000]

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

$v_{max}=a\omega \Rightarrow \omega =\frac{v_{max}}{a}=\frac{10}{4}$

Now , v = $\omega \sqrt{a^2-y^2}\Rightarrow v^2={\omega }^2(a^2-y^2)\Rightarrow y^2=a^2-\frac{v^2}{{\omega }^2}$

$\Rightarrow y=\sqrt{a^2-\frac{v^2}{{\omega }^2}}=\sqrt{4^2-\frac{5^2}{(\frac{10}{4}{)}^2}}=2\sqrt{3}$ cm