Question

A particle in S.H.M. has a period of $2$ seconds and amplitude of $10cm$. Calculate the acceleration when it is at $4cm$ from its positive extreme position.

Answer 1

$A=10cm$ $T=2s,x=4cm$

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

$a=-{\left(2\pi f\right)}^2\cdot x$

$={(2\times \frac{22}{7}\times f)}^2\cdot x$ (from positive extreme position)

$f=\frac{1}{T}=\frac{1}{2}$

$a={(2\times \frac{22}{7}\times \frac{1}{2})}^2\cdot x$

For $x=A-4$

$=10-4=6cm$

$=-{(2\times \frac{22}{7}\times \frac{1}{2})}^2\cdot 6$

$=\frac{22\times 22}{49}\cdot 6=\frac{2904}{49}$

$a=59.2m/s^2$.