91Ó°ÊÓ

A block with mass m = 0.4 kg is connected by a spring of relaxed length $l_0=0.15\mathrm{m}$ to a post at the centre of a low-friction table and the stretch length l = 0.28 m

The magnitude of an object's rate of change of position with time, or the magnitude of change of position per unit of time, is its speed; it is thus a scalar quantity.

Apply the concept of T for a spring mass system, first describe the spring constant in terms of the period T and the mass m.

$$\begin{gathered} T=2\mathrm{Ï€}\sqrt{\frac{\mathrm{m}}{\mathrm{k}}} \\ \mathrm{k}=\mathrm{m}{\left(\frac{2\mathrm{Ï€}}{\mathrm{T}}\right)}^2 \\ =\frac{4{\mathrm{Ï€}}^{2}m}{T^2} \end{gathered}$$

Continue writing Newton's second law for the mass along the direction of the spring while it is in circular motion.

Then replace k for the equation derived in the previous step.

The centripetal acceleration has been defined in terms of the tangential speed v and the stretched length l.

It's also worth noting that the stretch equals stretch length minus the relaxed length.

Finally, in this equation, solve for.

$$\begin{gathered} \underset{}{∑F=ks=m{a}_c} \\ \frac{4{\mathrm{π}}^2\mathrm{m}}{T^2}s=m{a}_c \\ {a}_c=\frac{v^2}{l} \\ \frac{4{\mathrm{π}}^2\mathrm{m}(\mathrm{l}-{\mathrm{l}}_{\mathrm{o}})}{T^2}=\frac{m{v}^2}{l} \\ v=\frac{2\mathrm{π}}{T}\sqrt{l(l-l_o)} \\ =\left(\frac{2\mathrm{π}}{0.6s}\right)\sqrt{(0.28m)(0.28m-0.15m)} \\ =2.0m/s \\ \mathrm{Therefore},\mathrm{the}\mathrm{speed}\mathrm{is}2.0\mathrm{m}/\mathrm{s}. \end{gathered}$$