91Ó°ÊÓ

Amplitude, $A=5cm$

Period, $T=1s$

If the masses and springs are the same, both vertical and horizontal spring-mass systems without friction oscillate equally around an equilibrium point. However, when it comes to vertical springs, keep in mind that gravity extends or compresses the spring beyond its natural length to get it to the equilibrium position.

Expression for the time period for the spring-mass system,

$T=\frac{2\mathrm{Ï€}}{\mathrm{Ó\neg }}...................................\left(1\right)$

Where is period, is the angular frequency.

$\mathrm{Ó\neg }=\sqrt{\frac{k}{m}}........................................\left(2\right)$

Where$k$ is the stiffness of spring,$m$ is the mass of the block.

Put the equation number 2 in equation (1), and we get,

$T=2\mathrm{Ï€}\sqrt{\frac{m}{k}}.......................................\left(3\right)$

Part a)

If we doubled the mass,

$M=2m$

Then from the equation (3),

$$\begin{gathered} T\text{'}=2\mathrm{π}\sqrt{\frac{2\mathrm{m}}{\mathrm{k}}} \\ \mathrm{T}\text{'}=\sqrt{2}×2\mathrm{π}\sqrt{\frac{\mathrm{m}}{\mathrm{k}}} \\ \mathrm{T}\text{'}=\sqrt{2}\mathrm{T} \end{gathered}$$

So, the time period increases by the root two times.

Part b)

The time period, if we doubled the stiffness of spring,

$K=2k$then from the equation (3)

$$\begin{gathered} T\text{'}=2\mathrm{π}\sqrt{\frac{\mathrm{m}}{2\mathrm{k}}} \\ \mathrm{T}\text{'}=\frac{1}{\sqrt{2}}×2\mathrm{π}\sqrt{\frac{\mathrm{m}}{\mathrm{k}}} \\ \mathrm{T}\text{'}=\frac{\mathrm{T}}{\sqrt{2}} \end{gathered}$$

So, the time period decreases by the root two times.

Part c)

From the above equation, it is independent of the length, so if we cut the original spring in half and use just one of the pieces. It is independent of time. So that time period will be the same.

Part d)

If we increased the amplitude of the original system to 10 cm, as the amplitude is independent of the time period, that time period would be the same.

Part e)