91Ó°ÊÓ

Given data can be listed below,

• Mass, $m=20g$

• Period, $T=1.2s$

Expression for the period of oscillation is given by

$\mathrm{Ï€}=2\frac{\sqrt{m}}{k}$

Substituting $1.2s$for $T$, and $20g$for $m$in the above equation

$$\begin{gathered} 1.2=2\mathrm{Ï€}\sqrt{\frac{\left(20g\right)\left(0.001kg\right)}{k\left(1g\right)}} \\ k=0.548N/m \end{gathered}$$

Part a)

Expression for the period of oscillation is given by

$T=2\sqrt{\frac{m}{k}}$

The value of mass is $2m$

Substituting $0.548N/m$for $k$, and $40g$for $m$in the above equation

role="math" localid="1657801494171" $$\begin{gathered} T=2\sqrt{\frac{\left(40g\right)\left(0.001kg\right)}{\left(0.548N/m\right)}} \\ =1.7s \end{gathered}$$

Part b)

When springs in parallel, then the equivalent spring stiffness is given by

$$\begin{gathered} k_e=k_1+k_2 \\ =0.548+0.548 \\ =1.096 \end{gathered}$$

Expression for the period of oscillation is given by

role="math" localid="1657801482565" $T=2\sqrt{\frac{m}{k}}$

Substituting $1.096N/m$for $k$, and $20g$for $m$in the above equation

role="math" localid="1657801458071" $$\begin{gathered} T=2\sqrt{\frac{\left(20g\right)\left(0.001kg\right)}{\left(1.096N/m\right)}} \\ =0.848s \end{gathered}$$

Part c)

Now we cut one spring into two equal lengths, and we hang one $20g$mass from this half spring, then the stiffness of the spring is

$k\text{'}=nk$

Where n is the number of parts,

$$\begin{gathered} k\text{'}=2×0.548 \\ =1.096N/m \end{gathered}$$

The period of oscillation is $0.848s$

Part d)

The period of oscillation of earth is the same as the period of oscillation measured on the moon. Because from the above equation of period of oscillation is depends on mass and stiffness.