Question

The velocities of a particle executing $S.H.M.$ are $10cm/s$ and $8cm/s$ at displacements $4cm$ and $5cm$ from mean position respectively. Calculate the time period of the particle.

Answer 1

Given when $y_1=4cm,v=-10c{m}^{-1}$

and when $y_2=5cm,v=8cm{s}^{-1}$

$T=?$

$\because$ Velocity in S.H.M.

$v={\omega }_0\sqrt{a^2-y^2}$

$\therefore v_1={\omega }_0\sqrt{a^2-y_1^2}$

or $v_1^2={\omega }_0^2(a^2-y_1^2).........\left(1\right)$

Similarly $v_2^2={\omega }_0^2\left(a^2{y}_2^2\right)...\left(2\right)$

Dividing eq (1) by eq (2) gives

$\frac{v_1^2}{v_2^2}=\frac{a^2-y_1^2}{a^2-y_2^2}$

$\therefore (\frac{10}{8}{)}^2=\frac{a^2-(4{)}^2}{a^2-(5{)}^2}$

or $\frac{25}{16}=\frac{a^2-16}{a^2-25}$

or $25a^2-25\times 25=12a^2-16\times 16$

$25a^2-16a^2=-256+625$

$9a^2=369$

or $3a=19.209$

$\therefore a=\frac{19.209}{3}=6.4cm$

From eq(1)

${\omega }_0^2=\frac{v_1^2}{a^2-y_1^2}=\frac{10\times 10}{41-16}=\frac{100}{25}$

$\therefore {\omega }_0=\frac{10}{5}=2$

or $\frac{2\pi }{T}=2\Rightarrow T=\pi$

$\therefore T=3.14s$