91Ó°ÊÓ

Given.

• $x=20.0cm=0.2m$
  
• $m=500g=0.5kg$
  

Required.

• $k$
• $T$
• $V_{max}$

$a)$At $x=20cm$the weight equalizes the restoring force and spring is in equilibrium so apply newton's first law to get a spring constant

$kx=mg⇒\left(1\right)$

Substitution in $\left(1\right)$yields

$k=\frac{mg}{x}=\frac{0.5×9.81}{0.2}=24.5N/m$

$b)$Now the time is given by

$\mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{T}=\sqrt{\frac{k}{m}}⇒\left(2\right)$

Substituting in $\left(2\right)$to get that

$T=2\mathrm{Ï€}\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}=2\mathrm{Ï€}\sqrt{\frac{0.5}{24.5}}=0.9\mathrm{s}$

The equilibrium position is $x=20cm$and the force acting on spring is zero so if we stretch spring another $10cm$the spring is no longer in equilibrium but it will be in simple harmonic motion and the velocity equation is

$v\left(t\right)=-v_{max}\sin \left(\mathrm{Ó\neg }t+{\mathrm{Ï•}}_0\right)⇒\left(3\right)$

Where $v_{max}$is given by

$v_{max}=\mathrm{Ó\neg }A⇒\left(4\right)$

Where:

• $A=10cm$
  

and $\mathrm{Ó\neg }$is given by substitution in $\left(2\right)$

$\mathrm{Ó\neg }=\sqrt{\frac{24.5}{0.5}}=7.0rad/s$

So substitution in $\left(4\right)$yields

$v_{max}=10×7=70cm/s$