Question

The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes $2cm,1m{s}^{-1}and10m{s}^{-2}$ at a certain instant. Find the amplitude and the time period of the motion.

• A. 4.9 cm, 0.28s
• B. 9.8 cm, 1s
• C. 4.9 cm, 2s
• D. 9.8 cm, 0.28s

Answer 1

The correct option is A

4.9 cm, 0.28s

Let's say the equation of SHM is

$x=Asin(\omega t+{\varphi }_0)$.....(1)

$\frac{dx}{dt}=v=A\omega cos(\omega t+{\varphi }_0)$....(2)

$\frac{d^{2}x}{d{t}^2}=a=-A{\omega }^{2}sin(\omega t+{\psi }_0)$....(3)

From equation (1) and (3)

$A=-{\omega }^{2}x$

Given magnitude of $acc=1000cm{s}^{-2}$

Displacement=1 cm

As the sign of acceleration and displacement is always opposite

$-1000=-{\omega }^2\times 2\Rightarrow \omega =\sqrt{500}$

$T=\frac{2\pi }{\omega }=\frac{2\pi }{\sqrt{500}}=0.28s$

Also from (1) and (3) we have,

$v=\omega \sqrt{A^2-x^2}\Rightarrow 100=\sqrt{500}\sqrt{A^2-2^2}$

$\Rightarrow A=4.9cm$