A block oscillating on a spring has a maximum speed of $20cm/s$. What will the block's maximum speed be if the total energy is doubled? Explain.

The expression for total energy of a block oscillating on a spring is,

$E_{total}=\frac{1}{2}k{A}^2$.......(1)

Here, $k$is spring constant and $A$is amplitude of oscillation.

Equate the total kinetic energy with total kinetic energy at the mean position.

$E_{kinetic}=\frac{1}{2}m{v}_{\max }^2$......(2)

Here, $m$ is mass of the block and $v_{max}$ is maximum speed.

Obtain a relation between maximum speed and total kinetic energy.

Equate the expression $\left(1\right)$and $\left(2\right)$

$\frac{1}{2}m{v}_{\max }^2=E_{total}$

$v_{\max }^2=\frac{2E_{total}}{m}$

$v_{\max }=\sqrt{\frac{2E_{total}}{m}}$

$v_{\max }\propto \sqrt{E_{total}}$

Calculate the block's maximum speed on doubling the total energy.

$\frac{v_{\max }^1}{v_{\max }}=\sqrt{\frac{{E^1}_{total}}{E_{total}}}$

$v_{\max }^1=\sqrt{\frac{{E^1}_{total}}{E_{total}}}\left(v_{\max }\right)$

Substitute $2$for $\frac{{E^1}_{total}}{E_{total}}$and $20cm/s$for $v_{max}$in the above equation

$v_{\max }^{\text{'}}=\sqrt{2}(20\mathrm{cm}/\mathrm{s})$

localid="1648224030351" $=28.3cm/s$.

Whenever the total energy is twice, the block's maximum speed is$28.3cm/s$.