Question

An oscillator consists of a block attached to a spring $(k=400N/m)$. At some time t, the position (measured from the systems equilibrium location), velocity and acceleration of the block are $x=0.100m,v=-15.0m/s$ , and $a=-90m/s^2$ . The amplitude of the motion and the mass of the block are

• A. 0.2 m, 0.84 kg
• B. 0.3 m, 0.76 kg
• C. 0.4 m, 0.54 kg
• D. 0.5 m, 0.44 kg

Answer 1

Answer: (D)

Given that, spring constant $K=400N/m$
Position $y=0.100m$
Velocity $V=-15.0m/s$
and acceleration $a=90m/s^2$
We know that $v=\omega \sqrt{A^2-y^2}\dots$. (i)
and $a=-{\omega }^{2}y$ ... (ii)
From Eq. (ii), $90={\omega }^2\times 0.1$
$\Rightarrow \omega =30$
How ${\omega }^2=\frac{k}{m}9\omega =\frac{400}{m}m=\frac{4}{9}=0.44kg$
From Eq (i), $15=\omega \sqrt{A^2-y^2}$
$225=900\left(A^2-y^2\right)$
$225=900A^2-900(0-1{)}^2$
$A^2=\frac{234}{900}A=\frac{15}{30}$
$=\frac{1}{2}=0.5m$