Question

A block (mass 5 kg) oscillates on a spring (spring constant 180 N/ m). At one moment, the block is 10 cm from its equilibrium position and is moving with a speed of 80 cm/s away from the equilibrium position. Determine the amplitude of this oscillation.

Answer 1

Let the amplitude of the oscillation is A.

Let the period of the oscillation is T.

Given that the mass of the block m=5 kg.

The spring constant k=180 N/m.

The speed of the block v=80 cm/s=0.80 m/s.

The position of the block x=10 cm=0.10 m.

Now we know the period of the oscillation

$T=2\pi \sqrt{\frac{m}{k}}$

Now we know the speed of the block at any point is

  $v=\pm \omega \sqrt{A^{2}-x^{2}}$

  $or\: \: \: \: \: \: v= \omega \sqrt{A^{2}-x^{2}}$

$or\: \: \: \: \: \: v=\frac{2\pi}{T} \times \sqrt{A^{2}-x^{2}}$

$or\: \: \: \: \: \: v=\frac{2\pi}{2\pi \sqrt{\frac{m}{k}}} \times \sqrt{A^{2}-x^{2}}$

  $or\: \: \: \: \: \: v=\sqrt{\frac{k}{m}} \times \sqrt{A^{2}-x^{2}}$

     $or\: \: \: \: \: \: v^{2}=\frac{k}{m}\times\left ( A^{2}-x^{2} \right )$

  $or\: \: \: \: \: \: A^{2}-x^{2} =\frac{mv^{2}}{k}$

  

or mv2 4² – 2 +0 k

     

my2 or A= +22 k

$or\: \: \: \: \: \: A =\sqrt{\frac{5\times (0.80)^{2}}{180}+0.10^{2}}$

$\therefore \: \: \: \: \: \: A =0.1666666667\approx 0.167\: \: \: \: m$

$\therefore \: \: \: \: \: \: A =16.7\: \: \: \: cm$

$\therefore$ The amplitude of the oscillation is 0.167 m or 16.7 cm approx.