Question

A particle performing S.H.M. having amplitude 'a' possesses velocity $\frac{(3{)}^{1/2}}{2}$ times the velocity at the mean position. The displacement of the particle shall be

• A. $a/2$
• B. $(3{)}^{1/2}\frac{a}{2}$
• C. $(\frac{a}{2}{)}^{1/2}$
• D. $(2{)}^{1/2}a$

Answer 1

Answer: (A)

$a/2$

In S.H.M the velocity of particle is

given by , $v^2=w^2(a^2-x^2)$

Where, a-amplitude of SHM

x-displacement of particle

At mean position $(x=0)$, let the velocity

of particle be $v$.

$\Rightarrow v^2=w^2{a}^2$

$\Rightarrow w^2=\frac{v^2}{a^2}$.....(1)

Now, let at a displacement $x$, the velocity

of particle is $\frac{\sqrt{3}}{2}v.$ We have to find $x$.

${\left(\frac{\sqrt{3}}{2}v\right)}^2=w^2(a^2-x^2)$

$\Rightarrow (\frac{\sqrt{3}}{2}v{)}^2=\frac{v^2}{a^2}(a^2-x^2)$ from (1)

$\Rightarrow \frac{3}{2}{v}^2=\frac{v^2}{a^2}(a^2-x^2)$

$\Rightarrow 3a^2=4(a^2-x^2)$

$\Rightarrow 3a^2=4a^2-4x^2$

$\Rightarrow 4x^2=a^2\Rightarrow x^2=\frac{a^2}{4}$

$\Rightarrow x=\frac{a}{2}$