Question

A particle executing simple harmonic motion with an amplitude $5$ cm and a time period $0.2$s. The velocity and acceleration of the particle when the displacement is $5$ cm is

• A. $0.5\pi m{s}^{-1},0m{s}^{-2}$
• B. $0.5m{s}^{-1},-5{\pi }^{2}m{s}^{-2}$
• C. $0m{s}^{-1},-5{\pi }^{2}m{s}^{-2}$
• D. $0.5\pi m{s}^{-1},-0.5{\pi }^{2}m{s}^{-2}$

Answer 1

The correct option is B $0.5m{s}^{-1},-5{\pi }^{2}m{s}^{-2}$

A particle executing simple harmonic motion with an amplitude$=5cm=\frac{5}{100}=0.05m$

Time period$=0.2s$

The velocity and acceleration of the particle when the displacement$=5cm$

$\omega =\frac{2\pi }{T}=\frac{2\pi }{0.2}=\frac{2\pi }{2}\times 10=10\pi rad/s$

And the displacement in $y$, then acceleration, $A={\omega }^{2}y$

And velocity$v=\omega \sqrt{r^2-y^2}$

$v=10\pi \sqrt{(0.05{)}^2-(0.05{)}^2}=0$

When $y=0,A=(10\pi {)}^2\times 0=0$

$v=10\pi \times \sqrt{(0.05{)}^2-0^2}=10\pi \times 0.05=0.5\pi m/s$