Question

It was found that a $\mathbf{20}\mathit{g}$mass hanging from a particular spring had an oscillation period of $\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{}\mathit{s}$. (a) When two $\mathbf{20}\mathbf{}\mathit{g}$masses are hung from this spring, what would you predict for the period in seconds? Explain briefly.

Figure 4.58

(b) When one $\mathbf{20}\mathbf{}\mathit{g}$mass is supported by two of these vertical, parallel springs (Figure 4.58), what would you predict for the period in seconds? Explain briefly. (c) Suppose that you cut one spring into two equal lengths, and you hang one $\mathbf{20}\mathbf{}\mathit{g}$mass from this half spring. What would you predict for the period in seconds? Explain briefly. (d) Suppose that you take a single (full-length) spring and a single $\mathbf{20}\mathbf{}\mathit{g}$mass to the Moon and watch the system oscillate vertically there. Will the period you observe on the Moon be longer, shorter, or the same as the period you measured on Earth? (The gravitational field strength on the Moon is about one-sixth that on the Earth.) Explain briefly.

Short answer

a) The period of oscillation is $1.7s$. for two masses.

b) The period of oscillation is $0.848s$for spring in parallel.

c) The period of oscillation is $0.848s$for cut one spring into two equal lengths.

d) The period of oscillation is $1.2s$, the period oscillation on the Moon be the same as the period you measured on Earth.

Step-by-step solution

Identification of given data

Given data can be listed below,

• Mass, $m=20g$

• Period, $T=1.2s$

Evaluating spring stiffness

Expression for the period of oscillation is given by

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

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

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

Evaluating the period of oscillation for two masses

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,

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

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.