Question

A particle vibrates in SHM along a straight line. Its greatest acceleration is $5{\pi }^2{\text{cm s}}^{-2}$ and when its distance from the equilibrium position is $4\text{cm}$, the velocity of the particle is $3{\pi }^2{\text{cm s}}^{-2}$. The amplitude and the period of oscillation of the vibrating particle(in SI units) are $a$ and $T$ respectively.
Report the sum of numerical values of $(a+T)$

Answer 1

Maximum acceleration of the particle is
${\text{(Acceleration)}}_{max}={\omega }^{2}a=5{\pi }^2...\left(1\right)$
Velocity at $y=4\text{cm}$ is,
$v=\omega \sqrt{a^2-y^2}=\omega \sqrt{a^2-4^2}=3\pi$
$\omega \sqrt{a^2-16}=3\pi .....\left(2\right)$
Squaring equation $\left(2\right)$ and dividing with equation $\left(1\right),$ we get
$\frac{{\omega }^2(a^2-16)}{{\omega }^{2}a}=\frac{9{\pi }^2}{5{\pi }^2}=\frac{9}{5}$
$\frac{a^2-16}{a}=\frac{9}{5}$
$5a^2-9a-80=0$
$5a^2-25a+16a-80=0$
$(5a+16)(a-5)=0$
$a=5\text{cm}$
From Equation (1),
${\omega }^2\times 5=5{\pi }^2\omega =\pi$
$\therefore T=2$ seconds

$\Rightarrow a+T=7$