A block with mass 0.4 kgis connected by a spring of relaxed length 0.15 mto a post at the centre of a low-friction table. You pull the block straight away from the post and release it, and you observe that the period of oscillation is 0.6 s. Next you stretch the spring to a length of 0.28mand give the block an initial speed vperpendicular to the spring, choosing vso that the motion is a circle with the post at the centre. What is this speed?

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.

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

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.

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