Question

A particle executing simple harmonic motion has angular frequency $6.28s^{-}1$ and amplitude $10cm$. Find $\left(x\right)$ the speed when the displacement is $6cm$, from the mean position, $\left(y\right)$ the speed at $t=\frac{1}{6}s$ assuming that the motion starts from rest at $t=0$.
(i) $25.1cm{s}^{-1}$
(ii) $54.4cm{s}^{-1}$
(iii) $50.2cm{s}^{-1}$
(iv) $27.2cm{s}^{-1}$

• A. $x$ - (ii); $y$ - (iv)
• B. $x$ - (iii); $y$ - (iv)
• C. $x$ - (i), $y$ - (iii)
• D. $x$ - (iii), $y$ - (ii)

Answer 1

The correct option is D

$x$ - (iii), $y$ - (ii)

$\left(x\right)v=\omega \sqrt{A^2-x^2}v=\left(6.28s^{-1}\right)\sqrt{(10cm{)}^2-(6cm{)}^2}v=50.2m/s$

$\left(y\right)Att=0$, the velocity is zero, i.e., the particle is at an extreme. The equation for displacement may be written as

$X=A\cos \omega t$

$Att=\frac{1}{6}s,v=-\left(0.1m\right)\left(6.28s^{-1}\right)\sin \left(\frac{6.28}{6}\right)$

$v=(-0.628m{s}^{-1})\sin \frac{\pi }{3}$

$v=54.4cm{s}^{-1}$.