Question

A block hangs in equilibrium from a vertical spring. When a second identical block is added, the original block sags by $5.0\mathrm{cm}$. What is the oscillation frequency of the two-block system?

Short answer

The two block system have oscillation frequency of$1.6\mathrm{Hz}$.

Step-by-step solution

Expression for ∆y

The spring force on the first block must balance the gravitational pull because it is in equilibrium. So

${\mathrm{F}}_{\mathrm{s}}={\mathrm{F}}_{\mathrm{g}}$

$\mathrm{k\Delta y}=\mathrm{mg}$

$\mathrm{\Delta y}=\frac{\mathrm{mg}}{\mathrm{k}}$

Calculation for km

Equilibrium is ,

$\mathrm{k}(\mathrm{\Delta y}+5.0\mathrm{cm})=(\mathrm{m}+\mathrm{m})\mathrm{g}$

$\mathrm{mg}+\mathrm{k}(0.05\mathrm{m})=2\mathrm{mg}$

$\mathrm{k}(0.05\mathrm{m})=\mathrm{mg}$

$\frac{\mathrm{k}}{\mathrm{m}}=\frac{\mathrm{g}}{0.05\mathrm{m}}$

$\frac{\mathrm{k}}{\mathrm{m}}=\frac{9.80\mathrm{m}/{\mathrm{s}}^2}{0.05\mathrm{m}}$

$=196{\mathrm{s}}^{-2}$

Calculation for oscillation frequency

Frequency is,

$=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{\mathrm{k}}{2\mathrm{m}}}$

$=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{196{\mathrm{s}}^{-2}}{2}}$

$=1.6\mathrm{Hz}$