Question

The maximum speed and acceleration of a particle executing simple harmonic motion are $10cm{s}^{-1}$ and $50cm{s}^{-2}$. Find the positions (s) of the particle when the speed is $8cm{s}^{-1}$.

Answer 1

$V_{max}=10cm/sec$

$\Rightarrow r\omega =10...\left(1\right)$

$\Rightarrow {\omega }^2=\frac{100}{r^2}$

$a_{max}={\omega }^{2}r=50cmse{c}^{-1}$

$\Rightarrow {\omega }^2=\frac{50}{r}=\frac{50}{r}...\left(2\right)$

$\therefore \frac{100}{r^2}=\frac{50}{r}$

$\Rightarrow r=2cm$

$\therefore \omega =\sqrt{\frac{100}{r^2}}=5se{c}^{-1}$

Again to find out the positions where the speed is 8 m/sec

$v^2-w^2(r^2-y^2)$

$\Rightarrow 64=25(4-y^2)$

$\Rightarrow \frac{64}{25}=4-y^2$

$\Rightarrow r-y^2=2.56$

$\Rightarrow y^2=1.44$

$\Rightarrow y=\sqrt{\left(1.44\right)}$

$\Rightarrow y=\pm 1.2cm$

from mean position